94
What Is Physics?
We have seen that part of physics is a study of motion, including accelerations,
which are changes in velocities. Physics is also a study of what can cause an object
to accelerate.That cause is a force, which is, loosely speaking, a push or pull on the
object. The force is said to act on the object to change its velocity. For example,
when a dragster accelerates, a force from the track acts on the rear tires to cause
the dragster’s acceleration.When a defensive guard knocks down a quarterback, a
force from the guard acts on the quarterback to cause the quarterback’s backward
acceleration. When a car slams into a telephone pole, a force on the car from the
CHAPTER 5
Force and Motion—I
5-1NEWTON’S FIRST AND SECOND LAWS
After reading this module, you should be able to . . .
5.01 Identify that a force is a vector quantity and thus has
both magnitude and direction and also components.
5.02 Given two or more forces acting on the same particle,
add the forces as vectors to get the net force.
5.03 Identify Newton’s first and second laws of motion.
5.04 Identify inertial reference frames.
5.05 Sketch a free-body diagram for an object, showing the
object as a particle and drawing the forces acting on it as
vectors with their tails anchored on the particle.
5.06 Apply the relationship (Newton’s second law) between
the net force on an object, the mass of the object, and the
acceleration produced by the net force.
5.07 Identify that only external forces on an object can cause
the object to accelerate.
●The velocity of an object can change (the object can accel-
erate) when the object is acted on by one or more forces
(pushes or pulls) from other objects. Newtonian mechanics
relates accelerations and forces.
●Forces are vector quantities. Their magnitudes are defined
in terms of the acceleration they would give the standard kilo-
gram. A force that accelerates that standard body by exactly
1 m/s2is defined to have a magnitude of 1 N. The direction of
a force is the direction of the acceleration it causes. Forces
are combined according to the rules of vector algebra. The
net force on a body is the vector sum of all the forces acting
on the body.
●If there is no net force on a body, the body remains at rest if
it is initially at rest or moves in a straight line at constant
speed if it is in motion.
●Reference frames in which Newtonian mechanics holds are
called inertial reference frames or inertial frames. Reference
frames in which Newtonian mechanics does not hold are
called noninertial reference frames or noninertial frames.
●The mass of a body is the characteristic of that body that
relates the body’s acceleration to the net force causing the
acceleration. Masses are scalar quantities.
●The net force on a body with mass mis related to the
body’s acceleration by
which may be written in the component versions
.
The second law indicates that in SI units
1 N 1 kgm/s2.
●A free-body diagram is a stripped-down diagram in which
only one body is considered. That body is represented by
either a sketch or a dot. The external forces on the body are
drawn, and a coordinate system is superimposed, oriented
so as to simplify the solution.
F
net,xmaxF
net,ymayand F
net,zmaz
Fnet
:ma
:,
a
:
Fnet
:
Key Ideas
Learning Objectives
95
5-1 NEWTON’S FIRST AND SECOND LAWS
pole causes the car to stop. Science, engineering, legal, and medical journals are
filled with articles about forces on objects, including people.
A Heads Up. Many students find this chapter to be more challenging than the
preceding ones. One reason is that we need to use vectors in setting up equations—
we cannot just sum some scalars. So, we need the vector rules from Chapter 3.
Another reason is that we shall see a lot of different arrangements: objects will
move along floors, ceilings, walls, and ramps. They will move upward on ropes
looped around pulleys or by sitting in ascending or descending elevators.
Sometimes, objects will even be tied together.
However, in spite of the variety of arrangements, we need only a single key
idea (Newton’s second law) to solve most of the homework problems. The pur-
pose of this chapter is for us to explore how we can apply that single key idea to
any given arrangement. The application will take experience—we need to solve
lots of problems, not just read words. So, let’s go through some of the words and
then get to the sample problems.
Newtonian Mechanics
The relation between a force and the acceleration it causes was first understood
by Isaac Newton (1642–1727) and is the subject of this chapter.The study of that
relation, as Newton presented it, is called Newtonian mechanics. We shall focus
on its three primary laws of motion.
Newtonian mechanics does not apply to all situations. If the speeds of the in-
teracting bodies are very large—an appreciable fraction of the speed of light—we
must replace Newtonian mechanics with Einstein’s special theory of relativity,
which holds at any speed, including those near the speed of light. If the interacting
bodies are on the scale of atomic structure (for example, they might be electrons
in an atom), we must replace Newtonian mechanics with quantum mechanics.
Physicists now view Newtonian mechanics as a special case of these two more
comprehensive theories. Still, it is a very important special case because it applies
to the motion of objects ranging in size from the very small (almost on the scale of
atomic structure) to astronomical (galaxies and clusters of galaxies).
Newton’s First Law
Before Newton formulated his mechanics, it was thought that some influence,
a “force,” was needed to keep a body moving at constant velocity. Similarly, a
body was thought to be in its “natural state” when it was at rest. For a body to
move with constant velocity, it seemingly had to be propelled in some way, by
a push or a pull. Otherwise, it would “naturally” stop moving.
These ideas were reasonable. If you send a puck sliding across a wooden
floor, it does indeed slow and then stop. If you want to make it move across the
floor with constant velocity, you have to continuously pull or push it.
Send a puck sliding over the ice of a skating rink, however, and it goes a lot
farther.You can imagine longer and more slippery surfaces, over which the puck
would slide farther and farther. In the limit you can think of a long, extremely
slippery surface (said to be a frictionless surface), over which the puck would
hardly slow. (We can in fact come close to this situation by sending a puck sliding
over a horizontal air table, across which it moves on a film of air.)
From these observations, we can conclude that a body will keep moving with
constant velocity if no force acts on it.That leads us to the first of Newton’s three
laws of motion:
Newton’s First Law: If no force acts on a body, the body’s velocity cannot
change; that is, the body cannot accelerate.
An inertial reference frame is one in which Newton’s laws hold.
96 CHAPTER 5 FORCE AND MOTION—I
In other words, if the body is at rest, it stays at rest. If it is moving, it continues to
move with the same velocity (same magnitude and same direction).
Force
Before we begin working problems with forces, we need to discuss several fea-
tures of forces, such as the force unit, the vector nature of forces, the combining of
forces, and the circumstances in which we can measure forces (without being
fooled by a fictitious force).
Unit. We can define the unit of force in terms of the acceleration a force
would give to the standard kilogram (Fig. 1-3), which has a mass defined to be ex-
actly 1 kg. Suppose we put that body on a horizontal, frictionless surface and pull
horizontally (Fig. 5-1) such that the body has an acceleration of 1 m/s2. Then we
can define our applied force as having a magnitude of 1 newton (abbreviated N).
If we then pulled with a force magnitude of 2 N, we would find that the accelera-
tion is 2 m/s2. Thus, the acceleration is proportional to the force. If the standard
body of 1 kg has an acceleration of magnitude a(in meters per second per sec-
ond), then the force (in newtons) producing the acceleration has a magnitude
equal to a.We now have a workable definition of the force unit.
Vectors. Force is a vector quantity and thus has not only magnitude but also
direction. So, if two or more forces act on a body, we find the net force (or result-
ant force) by adding them as vectors, following the rules of Chapter 3. A single
force that has the same magnitude and direction as the calculated net force
would then have the same effect as all the individual forces. This fact, called the
principle of superposition for forces, makes everyday forces reasonable and pre-
dictable. The world would indeed be strange and unpredictable if, say, you and a
friend each pulled on the standard body with a force of 1 N and somehow the net
pull was 14 N and the resulting acceleration was 14 m/s2.
In this book, forces are most often represented with a vector symbol such as
and a net force is represented with the vector symbol .As with other vectors,
a force or a net force can have components along coordinate axes.When forces act
only along a single axis, they are single-component forces. Then we can drop the
overhead arrows on the force symbols and just use signs to indicate the directions
of the forces along that axis.
The First Law. Instead of our previous wording, the more proper statement
of Newton’s First Law is in terms of a net force:
F
:
net
F
:
,
Figure 5-1 A force on the standard
kilogram gives that body an acceleration a
:.
F
:
a
F
Newton’s First Law: If no net force acts on a body , the body’s velocity
cannot change; that is, the body cannot accelerate.
(F
:
net 0)
There may be multiple forces acting on a body, but if their net force is zero, the
body cannot accelerate. So, if we happen to know that a body’s velocity is con-
stant, we can immediately say that the net force on it is zero.
Inertial Reference Frames
Newton’s first law is not true in all reference frames, but we can always find
reference frames in which it (as well as the rest of Newtonian mechanics) is true.
Such special frames are referred to as inertial reference frames, or simply inertial
frames.
For example, we can assume that the ground is an inertial frame provided we can
neglect Earth’s astronomical motions (such as its rotation).
97
5-1 NEWTON’S FIRST AND SECOND LAWS
That assumption works well if, say, a puck is sent sliding along a short strip
of frictionless ice—we would find that the puck’s motion obeys Newton’s laws.
However, suppose the puck is sent sliding along a long ice strip extending from
the north pole (Fig. 5-2a). If we view the puck from a stationary frame in space,
the puck moves south along a simple straight line because Earth’s rotation
around the north pole merely slides the ice beneath the puck. However, if we
view the puck from a point on the ground so that we rotate with Earth, the
puck’s path is not a simple straight line. Because the eastward speed of the
ground beneath the puck is greater the farther south the puck slides, from our
ground-based view the puck appears to be deflected westward (Fig. 5-2b).
However, this apparent deflection is caused not by a force as required by
Newton’s laws but by the fact that we see the puck from a rotating frame. In this
situation, the ground is a noninertial frame, and trying to explain the deflection
in terms of a force would lead us to a fictitious force.A more common example
of inventing such a nonexistent force can occur in a car that is rapidly increas-
ing in speed. You might claim that a force to the rear shoves you hard into the
seat back.
In this book we usually assume that the ground is an inertial frame and that
measured forces and accelerations are from this frame. If measurements are made
in, say, a vehicle that is accelerating relative to the ground, then the measurements
are being made in a noninertial frame and the results can be surprising.
Figure 5-2 (a) The path of a puck sliding
from the north pole as seen from a station-
ary point in space. Earth rotates to the east.
(b) The path of the puck as seen from the
ground.
N
S
E
W
(a)
(b)
Earth's rotation
causes an
apparent deflection.
Checkpoint 1
Which of the figure’s six arrangements correctly show the vector addition of forces
and to yield the third vector,which is meant to represent their net force ?F
:
net
F
:
2
F
:
1
(a) (c)(b)
F1F1F1
F1F1F1
F2
F2
F2
F2
F2F2
(d) (f)(e)
Mass
From everyday experience you already know that applying a given force to bod-
ies (say, a baseball and a bowling ball) results in different accelerations.The com-
mon explanation is correct: The object with the larger mass is accelerated less.
But we can be more precise. The acceleration is actually inversely related to the
mass (rather than, say, the square of the mass).
Let’s justify that inverse relationship. Suppose, as previously, we push on the
standard body (defined to have a mass of exactly 1 kg) with a force of magnitude
1 N. The body accelerates with a magnitude of 1 m/s2. Next we push on body X
with the same force and find that it accelerates at 0.25 m/s2. Let’s make the (cor-
rect) assumption that with the same force,
mX
m0
a0
aX
,
98 CHAPTER 5 FORCE AND MOTION—I
and thus
Defining the mass of Xin this way is useful only if the procedure is consis-
tent. Suppose we apply an 8.0 N force first to the standard body (getting an accel-
eration of 8.0 m/s2) and then to body X(getting an acceleration of 2.0 m/s2). We
would then calculate the mass of Xas
which means that our procedure is consistent and thus usable.
The results also suggest that mass is an intrinsic characteristic of a body—it
automatically comes with the existence of the body. Also, it is a scalar quantity.
However, the nagging question remains:What, exactly, is mass?
Since the word mass is used in everyday English, we should have some intu-
itive understanding of it, maybe something that we can physically sense. Is it
a body’s size, weight, or density? The answer is no, although those characteristics
are sometimes confused with mass. We can say only that the mass of a body is
the characteristic that relates a force on the body to the resulting acceleration. Mass
has no more familiar definition; you can have a physical sensation of mass only
when you try to accelerate a body, as in the kicking of a baseball or a bowling ball.
Newton’s Second Law
All the definitions, experiments, and observations we have discussed so far can be
summarized in one neat statement:
mXm0
a0
aX
(1.0 kg) 8.0 m/s2
2.0 m/s24.0 kg,
mXm0
a0
aX
(1.0 kg) 1.0 m/s2
0.25 m/s24.0 kg.
Newton’s Second Law: The net force on a body is equal to the product of the
body’s mass and its acceleration.
In equation form,
(Newton’s second law). (5-1)
Identify the Body. This simple equation is the key idea for nearly all the
homework problems in this chapter, but we must use it cautiously. First, we must
be certain about which body we are applying it to. Then must be the vector
sum of all the forces that act on that body. Only forces that act on that body are to
be included in the vector sum, not forces acting on other bodies that might be
involved in the given situation. For example, if you are in a rugby scrum, the net
force on you is the vector sum of all the pushes and pulls on your body. It does
not include any push or pull on another player from you or from anyone else.
Every time you work a force problem, your first step is to clearly state the body
to which you are applying Newton’s law.
Separate Axes. Like other vector equations, Eq. 5-1 is equivalent to three
component equations, one for each axis of an xyz coordinate system:
Fnet, xmax,Fnet, ymay, and Fnet, zmaz. (5-2)
Each of these equations relates the net force component along an axis to the
acceleration along that same axis. For example, the first equation tells us that
the sum of all the force components along the xaxis causes the xcomponent ax
of the body’s acceleration, but causes no acceleration in the yand zdirections.
Turned around, the acceleration component axis caused only by the sum of the
F
:
net
F
:
net ma
:
99
5-1 NEWTON’S FIRST AND SECOND LAWS
force components along the xaxis and is completely unrelated to force compo-
nents along another axis. In general,
The acceleration component along a given axis is caused only by the sum of the force
components along that same axis, and not by force components along any other axis.
Table 5-1 Units in Newton’s Second Law (Eqs. 5-1 and 5-2)
System Force Mass Acceleration
SI newton (N) kilogram (kg) m/s2
CGSadyne gram (g) cm/s2
Britishbpound (lb) slug ft/s2
a1 dyne 1gcm/s2.
b1lb1 slugft/s2.
Checkpoint 2
The figure here shows two horizontal forces acting
on a block on a frictionless floor.If a third horizon-
tal force also acts on the block, what are the magnitude and direction of when
the block is (a) stationary and (b) moving to the left with a constant speed of 5 m/s?
F
:
3
F
:
3
3 N 5 N
Forces in Equilibrium. Equation 5-1 tells us that if the net force on a body is
zero, the body’s acceleration . If the body is at rest, it stays at rest; if it is
moving, it continues to move at constant velocity. In such cases, any forces on the
body balance one another, and both the forces and the body are said to be in
equilibrium. Commonly, the forces are also said to cancel one another, but the
term “cancel” is tricky. It does not mean that the forces cease to exist (canceling
forces is not like canceling dinner reservations). The forces still act on the body
but cannot change the velocity.
Units. For SI units, Eq. 5-1 tells us that
1N(1 kg)(1 m/s2)1kgm/s2. (5-3)
Some force units in other systems of units are given in Table 5-1 and Appendix D.
Diagrams. To solve problems with Newton’s second law, we often draw a
free-body diagram in which the only body shown is the one for which we are sum-
ming forces. A sketch of the body itself is preferred by some teachers but, to save
space in these chapters, we shall usually represent the body with a dot.Also, each
force on the body is drawn as a vector arrow with its tail anchored on the body.A
coordinate system is usually included, and the acceleration of the body is some-
times shown with a vector arrow (labeled as an acceleration). This whole proce-
dure is designed to focus our attention on the body of interest.
a
:0
External Forces Only. Asystem consists of one or more bodies, and any
force on the bodies inside the system from bodies outside the system is called an
external force. If the bodies making up a system are rigidly connected to one an-
other, we can treat the system as one composite body, and the net force on it
is the vector sum of all external forces. (We do not include internal forces —that
is, forces between two bodies inside the system. Internal forces cannot accelerate
the system.) For example, a connected railroad engine and car form a system. If,
say, a tow line pulls on the front of the engine, the force due to the tow line acts on
the whole engine–car system. Just as for a single body, we can relate the net ex-
ternal force on a system to its acceleration with Newton’s second law, ,
where mis the total mass of the system.
F
:
net ma
:
F
:
net
100 CHAPTER 5 FORCE AND MOTION—I
Sample Problem 5.01 One- and two-dimensional forces, puck
Here are examples of how to use Newton’s second law for a
puck when one or two forces act on it. Parts A, B, and C of
Fig. 5-3 show three situations in which one or two forces act
on a puck that moves over frictionless ice along an xaxis, in
one-dimensional motion. The puck’s mass is m0.20 kg.
Forces and are directed along the axis and have
magnitudes F14.0 N and F22.0 N. Force is directedF
:
3
F
:
2
F
:
1
Figure 5-3 In three situations, forces act on a puck that moves
along an xaxis. Free-body diagrams are also shown.
F1
x
(a)
Puck x
A
(b)
F1
F1
F2
x
(c)
x
B
(d)
F1
F2
F2
x
x
(e)
C
(f)
θ
θ
F3
F2
F3
The horizontal force
causes a horizontal
acceleration.
This is a free-body
diagram.
These forces compete.
Their net force causes
a horizontal acceleration.
This is a free-body
diagram.
Only the horizontal
component of F3
competes with F2.
This is a free-body
diagram.
at angle u30and has magnitude F31.0 N. In each situ-
ation, what is the acceleration of the puck?
KEY IDEA
In each situation we can relate the acceleration to the net
force acting on the puck with Newton’s second law,
. However, because the motion is along only the x
axis, we can simplify each situation by writing the second
law for xcomponents only:
Fnet, xmax. (5-4)
The free-body diagrams for the three situations are also
given in Fig.5-3, with the puck represented by a dot.
Situation A: For Fig. 5-3b, where only one horizontal force
acts, Eq. 5-4 gives us
F1max,
which, with given data, yields
(Answer)
The positive answer indicates that the acceleration is in the
positive direction of the xaxis.
Situation B: In Fig. 5-3d, two horizontal forces act on the
puck, in the positive direction of xand in the negative
direction. Now Eq. 5-4 gives us
F1F2max,
which, with given data, yields
(Answer)
Thus, the net force accelerates the puck in the positive direc-
tion of the xaxis.
Situation C: In Fig. 5-3f, force is not directed along the
direction of the puck’s acceleration; only xcomponent F3,x
is. (Force is two-dimensional but the motion is only one-F
:
3
F
:
3
axF1F2
m4.0 N 2.0 N
0.20 kg 10 m/s2.
F
:
2
F
:
1
axF1
m4.0 N
0.20 kg 20 m/s2.
F
:
net ma
:
F
:
net
a
:
dimensional.) Thus, we write Eq. 5-4 as
F3,xF2max. (5-5)
From the figure, we see that F3,xF3cos u. Solving for the
acceleration and substituting for F3,xyield
(Answer)
Thus, the net force accelerates the puck in the negative di-
rection of the xaxis.
(1.0 N)(cos 30)2.0 N
0.20 kg 5.7 m/s2.
axF3,xF2
mF3 cos
F2
m
Additional examples, video, and practice available at WileyPLUS
101
5-1 NEWTON’S FIRST AND SECOND LAWS
x components: Along the xaxis we have
F3,xmaxF1,xF2,x
m(acos 50)F1cos(150)F2cos 90.
Then, substituting known data, we find
F3,x(2.0 kg)(3.0 m/s2) cos 50(10 N) cos(150)
(20 N) cos 90
12.5 N.
y components: Similarly, along the yaxis we find
F3,ymayF1,yF2,y
m(asin 50)F1sin(150)F2sin 90
(2.0 kg)(3.0 m/s2) sin 50(10 N) sin(150)
(20 N) sin 90
10.4 N.
Vector: In unit-vector notation,we can write
F3,xF3,y(12.5 N) (10.4 N)
(13 N) (10 N) . (Answer)
We can now use a vector-capable calculator to get the mag-
nitude and the angle of .We can also use Eq. 3-6 to obtain
the magnitude and the angle (from the positive direction of
the xaxis) as
and (Answer)
tan1F
3,y
F
3, x
40.
F
32F3,x
2F2
3,y16 N
F
:
3
j
ˆ
i
ˆ
j
ˆ
i
ˆ
j
ˆ
i
ˆ
F
:
3
Sample Problem 5.02 Two-dimensional forces, cookie tin
Here we find a missing force by using the acceleration. In
the overhead view of Fig. 5-4a, a 2.0 kg cookie tin is acceler-
ated at 3.0 m/s2in the direction shown by , over a friction-
less horizontal surface. The acceleration is caused by three
horizontal forces, only two of which are shown: of magni-
tude 10 N and of magnitude 20 N.What is the third force
in unit-vector notation and in magnitude-angle notation?
KEY IDEA
The net force on the tin is the sum of the three forces
and is related to the acceleration via Newton’s second law
.Thus,
, (5-6)
which gives us
(5-7)
Calculations: Because this is a two-dimensional problem,
we cannot find merely by substituting the magnitudes for
the vector quantities on the right side of Eq. 5-7. Instead, we
must vectorially add , (the reverse of ), and
(the reverse of ), as shown in Fig. 5-4b. This addition can
be done directly on a vector-capable calculator because we
know both magnitude and angle for all three vectors.
However, here we shall evaluate the right side of Eq. 5-7 in
terms of components, first along the xaxis and then along
the yaxis. Caution: Use only one axis at a time.
F
:
2
F
:
2
F
:
1
F
:
1
ma
:
F
:
3
F3
:ma
:F
:
1F2
:.
F
:
1F2
:F3
:ma
:
(F
:
net ma
:)
a
:
F
:
net
F
:
3
F
:
2
F
:
1
a
:
Additional examples, video, and practice available at WileyPLUS
Figure 5-4 (a) An overhead view of two of three horizontal forces that act on a cookie
tin, resulting in acceleration . is not shown. (b) An arrangement of vectors , ,
and to find force .F
:
3
F
:
2
F
:
1
ma
:
F
:
3
a
:
y
(a)
30°
x
y
(b)
x
F2
F3
F2
F1
a
a
50° m
–
F1
–
These are two
of the three
horizontal force
vectors.
This is the resulting
horizontal acceleration
vector.
We draw the product
of mass and acceleration
as a vector.
Then we can add the three
vectors to find the missing
third force vector.
Some Particular Forces
The Gravitational Force
Agravitational force on a body is a certain type of pull that is directed toward
a second body. In these early chapters, we do not discuss the nature of this force
and usually consider situations in which the second body is Earth.Thus, when we
speak of the gravitational force on a body, we usually mean a force that pulls
on it directly toward the center of Earth—that is, directly down toward the
ground.We shall assume that the ground is an inertial frame.
Free Fall. Suppose a body of mass mis in free fall with the free-fall accelera-
tion of magnitude g.Then, if we neglect the effects of the air, the only force acting
on the body is the gravitational force . We can relate this downward force andF
:
g
F
:
g
F
:
g
102 CHAPTER 5 FORCE AND MOTION—I
5-2 SOME PARTICULAR FORCES
After reading this module, you should be able to . . .
5.08 Determine the magnitude and direction of the gravita-
tional force acting on a body with a given mass, at a location
with a given free-fall acceleration.
5.09 Identify that the weight of a body is the magnitude of the
net force required to prevent the body from falling freely, as
measured from the reference frame of the ground.
5.10 Identify that a scale gives an object’s weight when the
measurement is done in an inertial frame but not in an ac-
celerating frame, where it gives an apparent weight.
5.11 Determine the magnitude and direction of the normal
force on an object when the object is pressed or pulled
onto a surface.
5.12 Identify that the force parallel to the surface is a frictional
force that appears when the object slides or attempts to
slide along the surface.
5.13 Identify that a tension force is said to pull at both ends of
a cord (or a cord-like object) when the cord is taut.
Learning Objectives
Key Ideas
●A gravitational force on a body is a pull by another body.
In most situations in this book, the other body is Earth or
some other astronomical body. For Earth, the force is directed
down toward the ground, which is assumed to be an inertial
frame. With that assumption, the magnitude of is
where mis the body’s mass and gis the magnitude of the
free-fall acceleration.
●The weight Wof a body is the magnitude of the upward force
needed to balance the gravitational force on the body. A body’s
weight is related to the body’s mass by
Wmg.
F
gmg,
Fg
:
Fg
:
●A normal force is the force on a body from a surface
against which the body presses. The normal force is always
perpendicular to the surface.
●A frictional force is the force on a body when the body
slides or attempts to slide along a surface. The force is always
parallel to the surface and directed so as to oppose the slid-
ing. On a frictionless surface, the frictional force is negligible.
●When a cord is under tension, each end of the cord pulls
on a body. The pull is directed along the cord, away from the
point of attachment to the body. For a massless cord (a cord
with negligible mass), the pulls at both ends of the cord have
the same magnitude T, even if the cord runs around a mass-
less, frictionless pulley (a pulley with negligible mass and
negligible friction on its axle to oppose its rotation).
f
:
FN
:
downward acceleration with Newton’s second law . We place a vertical
yaxis along the body’s path, with the positive direction upward. For this axis,
Newton’s second law can be written in the form Fnet,ymay, which, in our
situation, becomes
Fgm(g)
or Fgmg. (5-8)
In words, the magnitude of the gravitational force is equal to the product mg.
(F
:ma )
:
At Rest. This same gravitational force, with the same magnitude, still acts on
the body even when the body is not in free fall but is, say, at rest on a pool table or
moving across the table. (For the gravitational force to disappear, Earth would
have to disappear.)
We can write Newton’s second law for the gravitational force in these vector
forms:
Fgj
ˆmgj
ˆ(5-9)
where j
ˆis the unit vector that points upward along a yaxis, directly away from
the ground, and is the free-fall acceleration (written as a vector), directed
downward.
Weight
The weight Wof a body is the magnitude of the net force required to prevent the
body from falling freely, as measured by someone on the ground. For example, to
keep a ball at rest in your hand while you stand on the ground, you must provide
an upward force to balance the gravitational force on the ball from Earth.
Suppose the magnitude of the gravitational force is 2.0 N.Then the magnitude of
your upward force must be 2.0 N, and thus the weight Wof the ball is 2.0 N. We
also say that the ball weighs 2.0 N and speak about the ball weighing 2.0 N.
A ball with a weight of 3.0 N would require a greater force from you—
namely, a 3.0 N force—to keep it at rest.The reason is that the gravitational force
you must balance has a greater magnitude—namely, 3.0 N. We say that this sec-
ond ball is heavier than the first ball.
Now let us generalize the situation. Consider a body that has an acceleration
of zero relative to the ground, which we again assume to be an inertial frame.
Two forces act on the body: a downward gravitational force and a balancing
upward force of magnitude W.We can write Newton’s second law for a vertical y
axis, with the positive direction upward, as
Fnet,ymay.
In our situation, this becomes
WFgm(0) (5-10)
or WFg(weight, with ground as inertial frame). (5-11)
This equation tells us (assuming the ground is an inertial frame) that
F
:
g
a
:
g
:
mg
:,F
:
g
103
5-2 SOME PARTICULAR FORCES
The weight Wof a body is equal to the magnitude Fgof the gravitational force
on the body.
Substituting mg for Fgfrom Eq. 5-8, we find
Wmg (weight), (5-12)
which relates a body’s weight to its mass.
Weighing. To weigh a body means to measure its weight. One way to do this
is to place the body on one of the pans of an equal-arm balance (Fig. 5-5) and
then place reference bodies (whose masses are known) on the other pan until we
strike a balance (so that the gravitational forces on the two sides match). The
masses on the pans then match, and we know the mass of the body. If we know
the value of gfor the location of the balance, we can also find the weight of the
body with Eq. 5-12.
We can also weigh a body with a spring scale (Fig. 5-6). The body stretches
a spring, moving a pointer along a scale that has been calibrated and marked in
Figure 5-5 An equal-arm balance. When the
device is in balance, the gravitational force
on the body being weighed (on the leftF
:
gL
F
gL
= m
L
g F
gR
= m
R
g
m
R
m
L
Figure 5-6 A spring scale. The reading is
proportional to the weight of the object on
the pan, and the scale gives that weight if
marked in weight units. If, instead, it is
marked in mass units, the reading is the
object’s weight only if the value of gat the
location where the scale is being used is
the same as the value of gat the location
where the scale was calibrated.
Fg = mg
Scale marked
in either
weight or
mass units
pan) and the total gravitational force
on the reference bodies (on the right pan)
are equal. Thus, the mass mLof the body
being weighed is equal to the total mass
mRof the reference bodies.
F
:
gR
Figure 5-7 (a) A block resting on a table experiences a normal force perpendicular to
the tabletop. (b) The free-body diagram for the block.
F
:
N
Block
Normal force FN
(a)(b)
y
x
Block
Fg
Fg
FN
The normal force
is the force on
the block from the
supporting table.
The gravitational
force on the block
is due to Earth's
downward pull.
The forces
balance.
104 CHAPTER 5 FORCE AND MOTION—I
either mass or weight units. (Most bathroom scales in the United States work this
way and are marked in the force unit pounds.) If the scale is marked in
mass units, it is accurate only where the value of gis the same as where the scale
was calibrated.
The weight of a body must be measured when the body is not accelerating
vertically relative to the ground. For example, you can measure your weight on a
scale in your bathroom or on a fast train. However, if you repeat the measure-
ment with the scale in an accelerating elevator, the reading differs from your
weight because of the acceleration. Such a measurement is called an apparent
weight.
Caution: A body’s weight is not its mass. Weight is the magnitude of a force
and is related to mass by Eq. 5-12. If you move a body to a point where the value
of gis different, the body’s mass (an intrinsic property) is not different but the
weight is. For example, the weight of a bowling ball having a mass of 7.2 kg is 71 N
on Earth but only 12 N on the Moon. The mass is the same on Earth and Moon,
but the free-fall acceleration on the Moon is only 1.6 m/s2.
The Normal Force
If you stand on a mattress, Earth pulls you downward, but you remain stationary.
The reason is that the mattress, because it deforms downward due to you, pushes
up on you. Similarly, if you stand on a floor, it deforms (it is compressed, bent, or
buckled ever so slightly) and pushes up on you. Even a seemingly rigid concrete
floor does this (if it is not sitting directly on the ground, enough people on the
floor could break it).
The push on you from the mattress or floor is a normal force . The name
comes from the mathematical term normal, meaning perpendicular:The force on
you from, say, the floor is perpendicular to the floor.
F
:
N
When a body presses against a surface, the surface (even a seemingly rigid one)
deforms and pushes on the body with a normal force that is perpendicular to
the surface.
F
:
N
Figure 5-7ashows an example. A block of mass mpresses down on a table,
deforming it somewhat because of the gravitational force on the block. TheF
:
g
table pushes up on the block with normal force .The free-body diagram for the
block is given in Fig. 5-7b. Forces and are the only two forces on the block
and they are both vertical. Thus, for the block we can write Newton’s second law
for a positive-upward yaxis (Fnet, ymay) as
FNFgmay.
F
:
N
F
:
g
F
:
N
105
5-2 SOME PARTICULAR FORCES
From Eq. 5-8, we substitute mg for Fg, finding
Checkpoint 3
In Fig.5-7, is the magnitude of the normal force greater than, less than, or equal to
mg if the block and table are in an elevator moving upward (a) at constant speed and
(b) at increasing speed?
F
:
N
Friction
If we either slide or attempt to slide a body over a surface, the motion is resisted
by a bonding between the body and the surface. (We discuss this bonding more in
the next chapter.) The resistance is considered to be a single force called either
the frictional force or simply friction. This force is directed along the surface, op-
posite the direction of the intended motion (Fig.5-8). Sometimes, to simplify a sit-
uation, friction is assumed to be negligible (the surface, or even the body, is said
to be frictionless).
Tension
When a cord (or a rope, cable, or other such object) is attached to a body and
pulled taut, the cord pulls on the body with a force directed away from the
body and along the cord (Fig. 5-9a). The force is often called a tension force
because the cord is said to be in a state of tension (or to be under tension), which
means that it is being pulled taut.The tension in the cord is the magnitude Tof the
force on the body. For example, if the force on the body from the cord has magni-
tude T50 N, the tension in the cord is 50 N.
A cord is often said to be massless (meaning its mass is negligible compared
to the body’s mass) and unstretchable. The cord then exists only as a connection
between two bodies. It pulls on both bodies with the same force magnitude T,
T
:
f
:,
Figure 5-8 A frictional force opposes the
attempted slide of a body over a surface.
f
:
f
Direction of
attempted
slide
T
(a) (b) (c)
TT
T
T
T
The forces at the two ends of
the cord are equal in magnitude.
Figure 5-9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord
pulls on the body and the hand with force , even if the cord runs around a massless,
frictionless pulley as in (b) and (c).
T
:
FNmg may.
Then the magnitude of the normal force is
FNmg maym(gay) (5-13)
for any vertical acceleration ayof the table and block (they might be in an accel-
erating elevator). (Caution: We have already included the sign for gbut aycan be
positive or negative here.) If the table and block are not accelerating relative to
the ground, then ay0 and Eq. 5-13 yields
FNmg. (5-14)
Newton’s Third Law
Two bodies are said to interact when they push or pull on each other—that is,
when a force acts on each body due to the other body. For example, suppose you
position a book Bso it leans against a crate C(Fig. 5-10a). Then the book and
crate interact:There is a horizontal force on the book from the crate (or due
to the crate) and a horizontal force on the crate from the book (or due to the
book).This pair of forces is shown in Fig. 5-10b. Newton’s third law states that
F
:
CB
F
:
BC
Newton’s Third Law: When two bodies interact, the forces on the bodies from each
other are always equal in magnitude and opposite in direction.
For the book and crate,we can write this law as the scalar relation
FBC FCB (equal magnitudes)
or as the vector relation
(equal magnitudes and opposite directions), (5-15)
where the minus sign means that these two forces are in opposite directions. We
can call the forces between two interacting bodies a third-law force pair. When
F
:
BC F
:
CB
Figure 5-10 (a) Book Bleans against crate
C.(b) Forces (the force on the book
from the crate) and (the force on the
crate from the book) have the same mag-
nitude and are opposite in direction.
F
:
CB
F
:
BC
Crate CBook B
(a)
(b)
C
FCB
FBC
B
The force on B
due to Chas the same
magnitude as the
force on C due to B.
106 CHAPTER 5 FORCE AND MOTION—I
even if the bodies and the cord are accelerating and even if the cord runs around
amassless, frictionless pulley (Figs. 5-9band c). Such a pulley has negligible mass
compared to the bodies and negligible friction on its axle opposing its rotation. If
the cord wraps halfway around a pulley, as in Fig. 5-9c, the net force on the pulley
from the cord has the magnitude 2T.
Checkpoint 4
The suspended body in Fig.5-9cweighs 75 N. Is Tequal to,greater than, or less than
75 N when the body is moving upward (a) at constant speed, (b) at increasing speed,
and (c) at decreasing speed?
5-3 APPLYING NEWTON’S LAWS
After reading this module, you should be able to . . .
5.14 Identify Newton’s third law of motion and third-law force pairs.
5.15 For an object that moves vertically or on a horizontal or inclined
plane, apply Newton’s second law to a free-body diagram of the
object.
5.16 For an arrangement where a system of several objects
moves rigidly together, draw a free-body diagram and
apply Newton’s second law for the individual objects
and also for the system taken as a composite object.
●The net force on a body with mass mis related to the body’s
acceleration by
,
which may be written in the component versions
.F
net,xmaxF
net,ymayand F
net,zmaz
Fnet
:ma
:
a
:
Fnet
:
●If a force acts on body Bdue to body C, then there is
a force on body Cdue to body B:
The forces are equal in magnitude but opposite in directions.
FBC
:FCB
:.
FCB
:FBC
:
Learning Objectives
Key Ideas
107
Earth E
Table T
Cantaloupe C
(a)
(b)
(c)
FCE
(gravitational force)
FCT
(d)
FTC
Earth
Cantaloupe
FCT
(normal force from table)
FCE
FEC
These forces
just happen
to be balanced.
These are
third-law force
pairs.
So are these.
Figure 5-11 (a) A cantaloupe lies on a table that stands on Earth. (b) The forces on
the cantaloupe are and . (c) The third-law force pair for the cantaloupe–Earth
interaction. (d) The third-law force pair for the cantaloupe–table interaction.
F
:
CE
F
:
CT
5-3 APPLYING NEWTON’S LAWS
any two bodies interact in any situation, a third-law force pair is present. The
book and crate in Fig. 5-10aare stationary, but the third law would still hold if
they were moving and even if they were accelerating.
As another example, let us find the third-law force pairs involving the can-
taloupe in Fig. 5-11a, which lies on a table that stands on Earth. The cantaloupe
interacts with the table and with Earth (this time, there are three bodies whose
interactions we must sort out).
Let’s first focus on the forces acting on the cantaloupe (Fig. 5-11b). Force
is the normal force on the cantaloupe from the table, and force is the
gravitational force on the cantaloupe due to Earth. Are they a third-law force
pair? No, because they are forces on a single body, the cantaloupe, and not on
two interacting bodies.
To find a third-law pair, we must focus not on the cantaloupe but on the
interaction between the cantaloupe and one other body. In the cantaloupe–Earth
interaction (Fig. 5-11c), Earth pulls on the cantaloupe with a gravitational force
and the cantaloupe pulls on Earth with a gravitational force . Are these
forces a third-law force pair? Yes, because they are forces on two interacting bod-
ies, the force on each due to the other.Thus, by Newton’s third law,
(cantaloupe–Earth interaction).
Next, in the cantaloupe–table interaction, the force on the cantaloupe from
the table is and, conversely, the force on the table from the cantaloupe is
(Fig. 5-11d).These forces are also a third-law force pair,and so
(cantaloupe–table interaction).F
:
CT F
:
TC
F
:
TC
F
:
CT
F
:
CE F
:
EC
F
:
EC
F
:
CE
F
:
CE
F
:
CT
Checkpoint 5
Suppose that the cantaloupe and table of Fig.5-11 are in an elevator cab that begins to
accelerate upward. (a) Do the magnitudes of and increase, decrease,or stay
the same? (b) Are those two forces still equal in magnitude and opposite in direction?
(c) Do the magnitudes of and increase,decrease,or stay the same? (d) Are those
two forces still equal in magnitude and opposite in direction?
F
:
EC
F
:
CE
F
:
CT
F
:
TC
108 CHAPTER 5 FORCE AND MOTION—I
Applying Newton’s Laws
The rest of this chapter consists of sample problems. You should pore over
them, learning their procedures for attacking a problem. Especially important is
knowing how to translate a sketch of a situation into a free-body diagram with
appropriate axes, so that Newton’s laws can be applied.
Sample Problem 5.03 Block on table, block hanging
Figure 5-12 shows a block S(the sliding block) with mass
M3.3 kg. The block is free to move along a horizontal
frictionless surface and connected, by a cord that wraps over
a frictionless pulley, to a second block H(the hanging
block), with mass m2.1 kg.The cord and pulley have neg-
ligible masses compared to the blocks (they are “massless”).
The hanging block Hfalls as the sliding block Saccelerates
to the right. Find (a) the acceleration of block S, (b) the ac-
celeration of block H, and (c) the tension in the cord.
QWhat is this problem all about?
You are given two bodies—sliding block and hanging
block—but must also consider Earth, which pulls on both
bodies. (Without Earth, nothing would happen here.) A to-
tal of five forces act on the blocks, as shown in Fig.5-13:
1. The cord pulls to the right on sliding block Swith a force
of magnitude T.
2. The cord pulls upward on hanging block Hwith a force
of the same magnitude T. This upward force keeps block
Hfrom falling freely.
3. Earth pulls down on block Swith the gravitational force
which has a magnitude equal to Mg.
4. Earth pulls down on block Hwith the gravitational force
which has a magnitude equal to mg.
5. The table pushes up on block Swith a normal force .
There is another thing you should note.We assume that
the cord does not stretch, so that if block Hfalls 1 mm in a
F
:
N
F
:
gH,
F
:
gS,
certain time, block Smoves 1 mm to the right in that same
time. This means that the blocks move together and their
accelerations have the same magnitude a.
QHow do I classify this problem? Should it suggest a par-
ticular law of physics to me?
Yes. Forces, masses, and accelerations are involved, and
they should suggest Newton’s second law of motion,
.That is our starting key idea.
QIf I apply Newton’s second law to this problem, to which
body should I apply it?
We focus on two bodies, the sliding block and the hanging
block. Although they are extended objects (they are not
points), we can still treat each block as a particle because
every part of it moves in exactly the same way. A second key
idea is to apply Newton’s second law separately to each block.
QWhat about the pulley?
We cannot represent the pulley as a particle because
different parts of it move in different ways. When we dis-
cuss rotation, we shall deal with pulleys in detail.
Meanwhile, we eliminate the pulley from consideration by
assuming its mass to be negligible compared with the
masses of the two blocks. Its only function is to change the
cord’s orientation.
QOK. Now how do I apply to the sliding block?
Represent block Sas a particle of mass Mand draw all
the forces that act on it, as in Fig. 5-14a. This is the block’s
free-body diagram. Next, draw a set of axes. It makes sense
F
:
net ma
:
ma
:
F
:
net
Figure 5-12 A block Sof mass Mis connected to a block Hof mass
mby a cord that wraps over a pulley.
Sliding
block S
Hanging
block H
Frictionless
surface
M
m
F
gH
T
T
F
gS
Block H
Block S
m
M
F
N
Figure 5-13 The forces acting on the two blocks of Fig. 5-12.
109
5-3 APPLYING NEWTON’S LAWS
to draw the xaxis parallel to the table, in the direction in
which the block moves.
QThanks, but you still haven’t told me how to apply
to the sliding block.All you’ve done is explain
how to draw a free-body diagram.
You are right, and here’s the third key idea: The
expression is a vector equation, so we can write
it as three component equations:
Fnet,xMaxFnet,yMayFnet,zMaz(5-16)
in which Fnet,x,Fnet,y, and Fnet,zare the components of the net
force along the three axes. Now we apply each component
equation to its corresponding direction. Because block S
does not accelerate vertically, Fnet, yMaybecomes
FNFgS 0orFNFgS. (5-17)
Thus in the ydirection, the magnitude of the normal force is
equal to the magnitude of the gravitational force.
No force acts in the zdirection, which is perpendicular
to the page.
In the xdirection, there is only one force component,
which is T.Thus, Fnet, xMaxbecomes
TMa. (5-18)
This equation contains two unknowns, Tand a; so we cannot
yet solve it. Recall, however, that we have not said anything
about the hanging block.
QI agree. How do I apply to the hanging block?
We apply it just as we did for block S: Draw a free-body
diagram for block H,as in Fig. 5-14b.Then apply in
component form.This time, because the acceleration is along
the yaxis, we use the ypart of Eq. 5-16 (Fnet, ymay) to write
TFgH may. (5-19)
We can now substitute mg for FgH and afor ay(negative
F
:
net ma
:
F
:
net ma
:
F
:
net Ma
:
F
:
net ma
:
because block Haccelerates in the negative direction of the
yaxis).We find
Tmg ma. (5-20)
Now note that Eqs. 5-18 and 5-20 are simultaneous equa-
tions with the same two unknowns, Tand a. Subtracting
these equations eliminates T.Then solving for ayields
(5-21)
Substituting this result into Eq. 5-18 yields
(5-22)
Putting in the numbers gives, for these two quantities,
(Answer)
and
13 N. (Answer)
QThe problem is now solved, right?
That’s a fair question, but the problem is not really fin-
ished until we have examined the results to see whether
they make sense. (If you made these calculations on the job,
wouldn’t you want to see whether they made sense before
you turned them in?)
Look first at Eq. 5-21. Note that it is dimensionally
correct and that the acceleration awill always be less than g
(because of the cord, the hanging block is not in free fall).
Look now at Eq. 5-22, which we can rewrite in the form
(5-23)
In this form, it is easier to see that this equation is also
dimensionally correct, because both Tand mg have dimen-
sions of forces. Equation 5-23 also lets us see that the ten-
sion in the cord is always less than mg, and thus is always
less than the gravitational force on the hanging block.That is
a comforting thought because, if Twere greater than mg,
the hanging block would accelerate upward.
We can also check the results by studying special cases,
in which we can guess what the answers must be. A simple
example is to put g0, as if the experiment were carried out
in interstellar space. We know that in that case, the blocks
would not move from rest, there would be no forces on the
ends of the cord, and so there would be no tension in the
cord. Do the formulas predict this? Yes, they do. If you put
g0 in Eqs. 5-21 and 5-22, you find a0 and T0. Two
more special cases you might try are M0 and .m:
TM
Mmmg.
TMm
Mmg(3.3 kg)(2.1 kg)
3.3 kg 2.1 kg (9.8 m/s2)
3.8 m/s2
am
Mmg2.1 kg
3.3 kg 2.1 kg (9.8 m/s2)
TMm
Mmg.
am
Mmg.
Figure 5-14 (a) A free-body diagram for block Sof Fig. 5-12.
(b) A free-body diagram for block Hof Fig. 5-12.
M
Sliding
block S
x
y
m
Hanging
block H
x
y
FgH
T
FgS
T
a
a
(a) (b)
FN
Additional examples, video, and practice available at WileyPLUS
110 CHAPTER 5 FORCE AND MOTION—I
Sample Problem 5.04 Cord accelerates box up a ramp
Many students consider problems involving ramps (inclined
planes) to be especially hard. The difficulty is probably visual
because we work with (a) a tilted coordinate system and (b) the
components of the gravitational force, not the full force. Here is
a typical example with all the tilting and angles explained. (In
WileyPLUS, the figure is available as an animation with
voiceover.) In spite of the tilt, the key idea is to apply Newton’s
second law to the axis along which the motion occurs.
In Fig. 5-15a, a cord pulls a box of sea biscuits up along a
frictionless plane inclined at angle u30.0. The box has
mass m5.00 kg, and the force from the cord has magni-
tude T25.0 N. What is the box’s acceleration aalong the
inclined plane?
KEY IDEA
The acceleration along the plane is set by the force compo-
nents along the plane (not by force components perpendi-
cular to the plane), as expressed by Newton’s second law
(Eq. 5-1).
Calculations: We need to write Newton’s second law for
motion along an axis. Because the box moves along the in-
clined plane, placing an xaxis along the plane seems reason-
able (Fig. 5-15b). (There is nothing wrong with using our
usual coordinate system, but the expressions for compo-
nents would be a lot messier because of the misalignment of
the xaxis with the motion.)
After choosing a coordinate system, we draw a free-
body diagram with a dot representing the box (Fig. 5-15b).
Then we draw all the vectors for the forces acting on the box,
with the tails of the vectors anchored on the dot. (Drawing
the vectors willy-nilly on the diagram can easily lead to errors,
especially on exams, so always anchor the tails.)
Force from the cord is up the plane and has magni-
T
:
θ
y
x
FN
Fg
T
(b)
Cord
θ
(a)
The box accelerates.
Normal force
Cord's pull
Gravitational
force
x
T
mg sin
θ
(g)(h)
θ
mg cos
mg
θ
mg sin
θ
y
x
FN
(i)
mg cos
θ
The net of these
forces determines
the acceleration.
These forces
merely balance.
(e)
Fg
(d)(c)
90° −
θ
θ
90° −
θ
θθ
(f)
θ
This is a right
triangle.
Parallel
component of
Fg
This is also.
Hypotenuse
Adjacent leg
(use cos )
θ
Opposite leg
(use sin )
θ
Perpendicular
component of
Fg
Figure 5-15 (a) A box is pulled up a plane by a
cord. (b) The three forces acting on the
box: the cord’s force the gravitational force
and the normal force (c)–(i) Finding
the force components along the plane and
perpendicular to it. In WileyPLUS, this figure
is available as an animation with voiceover.
FN
:.Fg
:,
T
:
,
A
tude T25.0 N. The gravitational force is downward (ofFg
:
111
5-3 APPLYING NEWTON’S LAWS
course) and has magnitude mg (5.00 kg)(9.80 m/s2)49.0 N. plane and thus cannot affect the motion along the plane. (It
has no component along the plane to accelerate the box.)
We are now ready to write Newton’s second law for mo-
tion along the tilted xaxis:
The component axis the only component of the acceleration
(the box is not leaping up from the plane, which would be
strange, or descending into the plane, which would be even
stranger). So,let’s simply write afor the acceleration along the
plane. Because is in the positive xdirection and the compo-
nent mg sin uis in the negative xdirection,we next write
Tmg sin uma. (5-24)
Substituting data and solving for a, we find
a0.100 m/s2. (Answer)
The result is positive,indicating that the box accelerates up the
inclined plane, in the positive direction of the tilted xaxis. If
T
:
Fnet,xmax.
That direction means that only a component of the force is
along the plane, and only that component (not the full force)
affects the box’s acceleration along the plane. Thus, before we
can write Newton’s second law for motion along the xaxis, we
need to find an expression for that important component.
Figures 5-15cto hindicate the steps that lead to the ex-
pression. We start with the given angle of the plane and
work our way to a triangle of the force components (they
are the legs of the triangle and the full force is the hy-
potenuse). Figure 5-15cshows that the angle between the
ramp and is 90u. (Do you see a right triangle there?)Fg
:
Next, Figs. 5-15dto fshow and its components: One com-
ponent is parallel to the plane (that is the one we want) and
the other is perpendicular to the plane.
Because the perpendicular component is perpendicular,
the angle between it and must be u(Fig. 5-15d).The com-
ponent we want is the far leg of the component right trian-
gle. The magnitude of the hypotenuse is mg (the magnitude
of the gravitational force).Thus, the component we want has
magnitude mg sin u(Fig. 5-15g).
We have one more force to consider, the normal force
shown in Fig. 5-15b. However, it is perpendicular to theFN
:
Fg
:
Fg
:
Sample Problem 5.05 Reading a force graph
Here is an example of where you must dig information out
of a graph, not just read off a number. In Fig. 5-16a, two
forces are applied to a 4.00 kg block on a frictionless floor,
but only force is indicated. That force has a fixed magni-
tude but can be applied at an adjustable angle to the posi-
tive direction of the xaxis. Force is horizontal and fixed in
both magnitude and angle. Figure 5-16bgives the horizontal
acceleration axof the block for any given value of ufrom 0
to 90.What is the value of axfor u180?
KEY IDEAS
(1) The horizontal acceleration axdepends on the net hori-
zontal force Fnet, x, as given by Newton’s second law. (2) The
net horizontal force is the sum of the horizontal compo-
nents of forces and .
Calculations: The xcomponent of is F2because the vector
is horizontal. The xcomponent of is F1cos . Using these
expressions and a mass mof 4.00 kg, we can write Newton’s
second law ( m) for motion along the xaxis as
F1cos uF24.00ax. (5-25)
From this equation we see that when angle u90,F1cos u
is zero and F24.00ax. From the graph we see that the
a
:
F
:
net
F
:
1
F
:
2
F
:
2
F
:
1
F
:
2
F
:
1
Figure 5-16 (a) One of the two forces applied to a block is shown.
Its angle ucan be varied. (b) The block’s acceleration component
axversus u.
When F1 is horizontal,
the acceleration is
3.0 m/s2.
F1
x
θ
(a)
(b)
3
2
1
00° 90°
θ
ax (m/s2)
When F1 is vertical,
the acceleration is
0.50 m/s2.
corresponding acceleration is 0.50 m/s2. Thus, F22.00 N
and must be in the positive direction of the xaxis.F
:
2
we decreased the magnitude of enough to make a0, the
box would move up the plane at constant speed.And if we de-
crease the magnitude of even more, the acceleration would
be negative in spite of the cord’s pull.
T
:
T
:
Additional examples, video, and practice available at WileyPLUS
From Eq. 5-25, we find that when u0,
F1cos 02.00 4.00ax. (5-26)
From the graph we see that the corresponding acceleration
is 3.0 m/s2. From Eq. 5-26, we then find that F110 N.
Substituting F110 N, F22.00 N, and u180into
Eq. 5-25 leads to
ax2.00 m/s2. (Answer)
112 CHAPTER 5 FORCE AND MOTION—I
Sample Problem 5.06 Forces within an elevator cab
Although people would surely avoid getting into the ele-
vator with you, suppose that you weigh yourself while on
an elevator that is moving. Would you weigh more than,
less than, or the same as when the scale is on a stationary
floor?
In Fig. 5-17a, a passenger of mass m72.2 kg stands on
a platform scale in an elevator cab. We are concerned with
the scale readings when the cab is stationary and when it is
moving up or down.
(a) Find a general solution for the scale reading, whatever
the vertical motion of the cab.
KEY IDEAS
(1) The reading is equal to the magnitude of the normal force
on the passenger from the scale. The only other force act-
ing on the passenger is the gravitational force , as shown in
the free-body diagram of Fig. 5-17b. (2) We can relate the
forces on the passenger to his acceleration by using
Newton’s second law . However, recall that we
can use this law only in an inertial frame. If the cab acceler-
ates, then it is not an inertial frame. So we choose the ground
to be our inertial frame and make any measure of the passen-
ger’s acceleration relative to it.
Calculations: Because the two forces on the passenger and
his acceleration are all directed vertically, along the yaxis in
Fig. 5-17b, we can use Newton’s second law written for y
components (Fnet, ymay) to get
FNFgma
or FNFgma. (5-27)
(F
:
net ma
:)
a
:
F
:
g
F
:
N
FN
y
(b)(a)
Passenger
Fg
These forces
compete.
Their net force
causes a vertical
acceleration.
Figure 5-17 (a) A passenger stands on a platform scale that indi-
cates either his weight or his apparent weight. (b) The free-body
diagram for the passenger, showing the normal force on him
from the scale and the gravitational force .
F
:
g
F
:
N
This tells us that the scale reading, which is equal to normal
force magnitude FN, depends on the vertical acceleration.
Substituting mg for Fggives us
FNm(ga) (Answer) (5-28)
for any choice of acceleration a. If the acceleration is up-
ward, ais positive; if it is downward, ais negative.
(b) What does the scale read if the cab is stationary or
moving upward at a constant 0.50 m/s?
KEY IDEA
For any constant velocity (zero or otherwise), the accelera-
tion aof the passenger is zero.
Calculation: Substituting this and other known values into
Eq. 5-28, we find
FN(72.2 kg)(9.8 m/s20) 708 N.
(Answer)
This is the weight of the passenger and is equal to the mag-
nitude Fgof the gravitational force on him.
(c) What does the scale read if the cab accelerates upward at
3.20 m/s2and downward at 3.20 m/s2?
Calculations: For a3.20 m/s2,Eq. 5-28 gives
FN(72.2 kg)(9.8 m/s23.20 m/s2)
939 N, (Answer)
and for a3.20 m/s2, it gives
FN(72.2 kg)(9.8 m/s23.20 m/s2)
477 N. (Answer)
For an upward acceleration (either the cab’s upward
speed is increasing or its downward speed is decreasing),
the scale reading is greater than the passenger’s weight.
That reading is a measurement of an apparent weight, be-
cause it is made in a noninertial frame. For a downward
acceleration (either decreasing upward speed or increas-
ing downward speed), the scale reading is less than the
passenger’s weight.
(d) During the upward acceleration in part (c), what is the
magnitude Fnet of the net force on the passenger, and what is
the magnitude ap,cab of his acceleration as measured in the
frame of the cab? Does ?
Calculation: The magnitude Fgof the gravitational force on
the passenger does not depend on the motion of the passen-
ger or the cab; so, from part (b), Fgis 708 N. From part (c), the
magnitude FNof the normal force on the passenger during
F
:
net ma
:
p,cab
113
5-3 APPLYING NEWTON’S LAWS
Dead-End Solution: Let us now include force by writ-
ing,again for the xaxis,
Fapp FAB mAa.
(We use the minus sign to include the direction of .)
Because FAB is a second unknown, we cannot solve this
equation for a.
Successful Solution: Because of the direction in which force
is applied,the two blocks form a rigidly connected system.
We can relate the net force on the system to the acceleration of
the system with Newton’s second law. Here, once again for the
xaxis, we can write that law as
Fapp (mAmB)a,
where now we properly apply to the system with
total mass mAmB. Solving for aand substituting known
values, we find
(Answer)
Thus, the acceleration of the system and of each block is in the
positive direction of the xaxis and has the magnitude 2.0 m/s2.
(b) What is the (horizontal) force on block Bfrom
block A(Fig. 5-18c)?
KEY IDEA
We can relate the net force on block Bto the block’s accel-
eration with Newton’s second law.
Calculation: Here we can write that law, still for compo-
nents along the xaxis, as
FBA mBa,
which, with known values, gives
FBA (6.0 kg)(2.0 m/s2)12 N. (Answer)
Thus, force is in the positive direction of the xaxis and
has a magnitude of 12 N.
F
:
BA
F
:
BA
aFapp
mAmB
20 N
4.0 kg 6.0 kg 2.0 m/s2.
F
:
app
F
:
app
F
:
AB
F
:
AB
Sample Problem 5.07 Acceleration of block pushing on block
Some homework problems involve objects that move to-
gether, because they are either shoved together or tied to-
gether. Here is an example in which you apply Newton’s
second law to the composite of two blocks and then to the
individual blocks.
In Fig. 5-18a, a constant horizontal force of magni-F
:
app
tude 20 N is applied to block Aof mass mA4.0 kg, which
pushes against block Bof mass mB6.0 kg.The blocks slide
over a frictionless surface, along an xaxis.
(a) What is the acceleration of the blocks?
Serious Error: Because force is applied directly
to block A, we use Newton’s second law to relate that
force to the acceleration of block A. Because the motion
is along the xaxis, we use that law for xcomponents
(Fnet, xmax), writing it as
Fapp mAa.
However, this is seriously wrong because is not the
only horizontal force acting on block A. There is also the
force from block B(Fig. 5-18b).F
:
AB
F
:
app
a
:
F
:
app
Figure 5-18 (a) A constant horizontal force is applied to block
A, which pushes against block B.(b) Two horizontal forces act on
block A.(c) Only one horizontal force acts on block B.
F
:
app
FBA
(c)
x
B
(a)
x
A
B
Fapp
(b)
x
AFAB
Fapp
This force causes the
acceleration of the full
two-block system.
This is the only force
causing the acceleration
of block B.
These are the two forces
acting on just block A.
Their net force causes
its acceleration.
Additional examples, video, and practice available at WileyPLUS
the upward acceleration is the 939 N reading on the scale.Thus,
the net force on the passenger is
Fnet FNFg939 N 708 N 231 N, (Answer)
during the upward acceleration. However, his acceleration
ap,cab relative to the frame of the cab is zero. Thus, in the non-
inertial frame of the accelerating cab, Fnet is not equal to
map,cab, and Newton’s second law does not hold.
1Figure 5-19 gives the free-body diagram for four situations in
which an object is pulled by several forces across a frictionless
floor, as seen from overhead. In which situations does the accel-
eration of the object have (a) an xcomponent and (b) a ycom-a
:
Questions
xx
yy
7 N
3 N
2 N
4 N 4 N
2 N
2 N
6 N
5 N
3 N
(1) (2)
x
x
yy
3 N
3 N
3 N
4 N
4 N
5 N
6 N
2 N
5 N
5 N
4 N
(3) (4)
Figure 5-19 Question 1.
114 CHAPTER 5 FORCE AND MOTION—I
Newtonian Mechanics The velocity of an object can change
(the object can accelerate) when the object is acted on by one or
more forces (pushes or pulls) from other objects. Newtonian me-
chanics relates accelerations and forces.
Force Forces are vector quantities. Their magnitudes are de-
fined in terms of the acceleration they would give the standard
kilogram. A force that accelerates that standard body by exactly
1 m/s2is defined to have a magnitude of 1 N. The direction of a
force is the direction of the acceleration it causes. Forces are com-
bined according to the rules of vector algebra. The net force on a
body is the vector sum of all the forces acting on the body.
Newton’s First Law If there is no net force on a body, the
body remains at rest if it is initially at rest or moves in a straight
line at constant speed if it is in motion.
Inertial Reference Frames Reference frames in which
Newtonian mechanics holds are called inertial reference frames or
inertial frames. Reference frames in which Newtonian mechanics
does not hold are called noninertial reference frames or noniner-
tial frames.
Mass The mass of a body is the characteristic of that body that
relates the body’s acceleration to the net force causing the acceler-
ation. Masses are scalar quantities.
Newton’s Second Law The net force on a body with
mass mis related to the body’s acceleration by
(5-1)
which may be written in the component versions
Fnet, xmaxFnet, ymayand Fnet, zmaz. (5-2)
The second law indicates that in SI units
1N1kgm/s2. (5-3)
F
:
net ma
:,
a
:
F
:
net
Review & Summary
Afree-body diagram is a stripped-down diagram in which only
one body is considered.That body is represented by either a sketch or
a dot. The external forces on the body are drawn, and a coordinate
system is superimposed,oriented so as to simplify the solution.
Some Particular Forces Agravitational force on a body
is a pull by another body. In most situations in this book, the other
body is Earth or some other astronomical body. For Earth, the
force is directed down toward the ground, which is assumed to be
an inertial frame.With that assumption,the magnitude of is
Fgmg, (5-8)
where mis the body’s mass and gis the magnitude of the free-fall
acceleration.
The weight Wof a body is the magnitude of the upward force
needed to balance the gravitational force on the body. A body’s
weight is related to the body’s mass by
Wmg. (5-12)
Anormal force is the force on a body from a surface
against which the body presses. The normal force is always perpen-
dicular to the surface.
Africtional force is the force on a body when the body
slides or attempts to slide along a surface. The force is always par-
allel to the surface and directed so as to oppose the sliding. On a
frictionless surface, the frictional force is negligible.
When a cord is under tension, each end of the cord pulls on a
body.The pull is directed along the cord,away from the point of at-
tachment to the body. For a massless cord (a cord with negligible
mass), the pulls at both ends of the cord have the same magnitude
T, even if the cord runs around a massless, frictionless pulley (a pul-
ley with negligible mass and negligible friction on its axle to op-
pose its rotation).
Newton’s Third Law If a force acts on body Bdue to
body C, then there is a force on body Cdue to body B:
F
:
BC F
:
CB.
F
:
CB
F
:
BC
f
:
F
:
N
F
:
g
F
:
g
ponent? (c) In each situation, give the direction of by naming
either a quadrant or a direction along an axis. (Don’t reach for
the calculator because this can be answered with a few mental
calculations.)
a
:
115
QUESTIONS
7July 17, 1981, Kansas City: The newly opened Hyatt
Regency is packed with people listening and dancing to a band
playing favorites from the 1940s. Many of the people are crowded
onto the walkways that hang like bridges across the wide atrium.
Suddenly two of the walkways collapse, falling onto the merrymak-
ers on the main floor.
The walkways were suspended one above another on vertical
rods and held in place by nuts threaded onto the rods. In the origi-
nal design, only two long rods were to be used, each extending
through all three walkways (Fig. 5-24a). If each walkway and the
merrymakers on it have a combined mass of M, what is the total
mass supported by the threads and two nuts on (a) the lowest
walkway and (b) the highest walkway?
Apparently someone responsible for the actual construction
realized that threading nuts on a rod is impossible except at the
ends, so the design was changed: Instead, six rods were used, each
connecting two walkways (Fig. 5-24b). What now is the total mass
supported by the threads and two nuts on (c) the lowest walkway,
(d) the upper side of the highest walkway, and (e) the lower side of
the highest walkway? It was this design that failed on that tragic
night—a simple engineering error.
2Two horizontal forces,
pull a banana split across a friction-
less lunch counter. Without using a
calculator, determine which of the
vectors in the free-body diagram of
Fig. 5-20 best represent (a) and
(b) . What is the net-force compo-
nent along (c) the xaxis and (d) the y
axis? Into which quadrants do (e) the
net-force vector and (f) the split’s ac-
celeration vector point?
3In Fig. 5-21, forces and
are applied to a lunchbox as it
slides at constant velocity over a
frictionless floor. We are to de-
crease angle uwithout changing the
magnitude of . For constant ve-
locity, should we increase, decrease,
or maintain the magnitude of ?
4At time t0, constant begins
to act on a rock moving through
deep space in the +xdirection. (a)
For time t0, which are possible functions x(t) for the rock’s posi-
tion: (1) x4t3, (2) x4t26t3, (3) x4t26t3? (b)
For which function is directed opposite the rock’s initial direction
of motion?
5Figure 5-22 shows overhead views of four situations in which
forces act on a block that lies on a frictionless floor. If the force
magnitudes are chosen properly, in which situations is it possible
that the block is (a) stationary and (b) moving with a constant
velocity?
F
:
F
:
F
:
2
F
:
1
F
:
2
F
:
1
F
:
2
F
:
1
F
:
1(3 N)i
ˆ(4 N)j
ˆ and F
:
2(1 N)i
ˆ(2 N)j
ˆ
Figure 5-21 Question 3.
y
x
58
4
32
67
1
Figure 5-20 Question 2.
θ
F2
F1
Figure 5-22 Question 5.
(1) F2
F1
F3
(3) (4)
(2) F1
F1
F1
F3
F2
F2
F2
6 N 3 N
(a)
60 N
5
8 N
(b)
15 N 13 N
(
c
)
25 N
20 N
43 N
(
d
)
Figure 5-23 Question 6.
6Figure 5-23 shows the same breadbox in four situations where
horizontal forces are applied. Rank the situations according to the
magnitude of the box’s acceleration,greatest first.
Rods
Nuts
Walkways
(a) (b)
Figure 5-24 Question 7.
8Figure 5-25 gives three graphs of velocity component vx(t) and
three graphs of velocity component vy(t). The graphs are not to
scale. Which vx(t) graph and which vy(t) graph best correspond to
each of the four situations in Question 1 and Fig.5-19?
Figure 5-25 Question 8.
vx
t
(a)
vx
t
(b)
vx
t
(c)
vy
(d)
vy
t
(e)
vy
t
(f)
t
••5 Three astronauts, propelled
by jet backpacks, push and guide a
120 kg asteroid toward a processing
dock, exerting the forces shown in
Fig. 5-29, with F132 N, F255 N,
F341 N, u130, and u360.
What is the asteroid’s acceleration
(a) in unit-vector notation and as (b) a magnitude and (c) a direc-
tion relative to the positive direction of the xaxis?
••6 In a two-dimensional tug-of-
war, Alex, Betty, and Charles pull
horizontally on an automobile tire at
the angles shown in the overhead
view of Fig. 5-30. The tire remains
stationary in spite of the three pulls.
Alex pulls with force of magni-
tude 220 N, and Charles pulls with
force of magnitude 170 N. Note
that the direction of is not given.
What is the magnitude of Betty’s
force
••7 There are two forces on the
2.00 kg box in the overhead view of
Fig. 5-31, but only one is shown. For
F1 20.0 N, a 12.0 m/s2,and u30.0,
find the second force (a) in unit-vector
notation and as (b) a magnitude and
(c) an angle relative to the positive di-
rection of the xaxis.
••8 A 2.00 kg object is subjected to
three forces that give it an acceleration
.If
two of the three forces are
and
find the third force.
••9 A 0.340 kg particle moves in an xy plane according
to x(t)15.00 2.00t4.00t3and y(t)25.00 7.00t9.00t2,
with xand yin meters and tin seconds. At t0.700 s, what are
(12.0 N)i
ˆ(8.00 N)j
ˆ,
F2
:F1
:(30.0 N)i
ˆ(16.0 N)j
ˆ
a
:(8.00 m/s2)i
ˆ(6.00 m/s2)j
ˆ
SSM
F
:
B?
F
:
C
F
:
C
F
:
A
Module 5-1 Newton’s First and Second Laws
•1 Only two horizontal forces act on a 3.0 kg body that can move
over a frictionless floor. One force is 9.0 N, acting due east, and the
other is 8.0 N, acting 62north of west. What is the magnitude of
the body’s acceleration?
•2 Two horizontal forces act on a 2.0 kg chopping block that can
slide over a frictionless kitchen counter, which lies in an xy plane.
One force is Find the acceleration of the
chopping block in unit-vector notation when the other force is
(a) (b)
and (c) .
•3 If the 1 kg standard body has an acceleration of 2.00 m/s2at
20.0to the positive direction of an xaxis, what are (a) the xcom-
ponent and (b) the ycomponent of the net force acting on the
body, and (c) what is the net force in unit-vector notation?
••4 While two forces act on it, a
particle is to move at the constant
velocity One
of the forces is
What is the other force?(6 N)j
ˆ.
F1
:(2 N)i
ˆ
v
:(3 m/s)i
ˆ(4 m/s)j
ˆ.
(4.0 N)j
ˆ
F
:
2(3.0 N)i
ˆ
(3.0 N)i
ˆ(4.0 N)j
ˆ,F
:
2F
:
2(3.0 N)i
ˆ(4.0 N)j
ˆ,
F
:
1(3.0 N)i
ˆ(4.0 N)j
ˆ.
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
x
y
F1
F2
F3
1
θ
3
θ
Figure 5-29 Problem 5.
Figure 5-30 Problem 6.
Alex
Charles
Betty
137°
x
y
θ
F1
a
Figure 5-31 Problem 7.
116 CHAPTER 5 FORCE AND MOTION—I
9Figure 5-26 shows a train of four blocks being pulled across a
frictionless floor by force . What total mass is accelerated to the
right by (a) force , (b) cord 3, and (c) cord 1? (d) Rank the blocks
according to their accelerations, greatest first. (e) Rank the cords
according to their tension, greatest first.
F
:F
:on block 3 from block 2? (d) Rank the blocks according to
their acceleration magnitudes, greatest first.(e) Rank forces , ,
and according to magnitude, greatest first.
11 A vertical force is applied to a block of mass mthat lies on
a floor.What happens to the magnitude of the normal force on
the block from the floor as magnitude Fis increased from zero if
force is (a) downward and (b) upward?
12 Figure 5-28 shows four choices for the direction of a force of
magnitude Fto be applied to a block
on an inclined plane. The directions
are either horizontal or vertical.
(For choice b, the force is not
enough to lift the block off the
plane.) Rank the choices according
to the magnitude of the normal
force acting on the block from the
plane, greatest first.
F
:
F
:
N
F
:
F
:
32
F
:
21
F
:
F
:
32
10 Figure 5-27 shows three blocks
being pushed across a frictionless
floor by horizontal force .What to-
tal mass is accelerated to the right
by (a) force , (b) force on
block 2 from block 1, and (c) force
F
:
21
F
:
F
:
Cord
1
Cord
2
Cord
3
10 kg 3 kg 5 kg 2 kg F
Figure 5-26 Question 9.
30°
a
c
b
d
Figure 5-28 Question 12.
21
2 kg
10 kg
5 kg
3
F
Figure 5-27 Question 10.
Module 5-2 Some Particular Forces
•13 Figure 5-33 shows an arrangement in
which four disks are suspended by cords. The
longer, top cord loops over a frictionless pul-
ley and pulls with a force of magnitude 98 N
on the wall to which it is attached.The tensions
in the three shorter cords are T158.8 N,
T249.0 N, and T39.8 N. What are the
masses of (a) disk A, (b) disk B, (c) disk C,
and (d) disk D?
•14 A block with a weight of 3.0 N is at
rest on a horizontal surface. A 1.0 N upward
force is applied to the block by means of an
attached vertical string. What are the (a)
magnitude and (b) direction of the force of
the block on the horizontal surface?
•15 (a) An 11.0 kg salami is supported by a cord that runs to
a spring scale, which is supported by a cord hung from the ceiling
(Fig. 5-34a).What is the reading on the scale, which is marked in SI
weight units? (This is a way to measure weight by a deli owner.) (b)
In Fig. 5-34bthe salami is supported by a cord that runs around a
pulley and to a scale.The opposite end of the scale is attached by a
cord to a wall.What is the reading on the scale? (This is the way by
a physics major.) (c) In Fig. 5-34cthe wall has been replaced with a
second 11.0 kg salami, and the assembly is stationary. What is the
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117
PROBLEMS
(a) the magnitude and (b) the angle (relative to the positive direc-
tion of the xaxis) of the net force on the particle, and (c) what is
the angle of the particle’s direction of travel?
••10 A 0.150 kg particle moves along an xaxis according
vx (m/s)
t (s)
4
2
01 2 3
–2
–4
Figure 5-32 Problem 12.
A
B
C
D
T
1
T
2
T
3
Figure 5-33
Problem 13.
reading on the scale? (This is the way by a deli owner who was
once a physics major.)
S
A
L
A
M
I
G
E
N
O
A
S
A
L
A
M
I
G
E
N
O
A
S
A
L
A
M
I
G
E
N
O
A
S
A
L
A
M
I
G
E
N
O
A
Spring scale
Spring scale
Spring
scale
(b)
(c)
(a)
Figure 5-34 Problem 15.
••16 Some insects can walk below
a thin rod (such as a twig) by hang-
ing from it. Suppose that such an in-
sect has mass mand hangs from a
horizontal rod as shown in Fig. 5-35,
with angle u40. Its six legs are all
under the same tension, and the leg
sections nearest the body are hori-
zontal. (a) What is the ratio of the
tension in each tibia (forepart of a leg) to the insect’s weight? (b) If
the insect straightens out its legs somewhat, does the tension in each
tibia increase, decrease,or stay the same?
Module 5-3 Applying
Newton’s Laws
•17 In Fig. 5-36,
let the mass of the block be
8.5 kg and the angle be 30.
Find (a) the tension in the cord
and (b) the normal force acting
on the block. (c) If the cord is
cut, find the magnitude of the re-
sulting acceleration of the block.
•18 In April 1974, John
Massis of Belgium managed to
move two passenger railroad
cars. He did so by clamping his teeth down on a bit that was at-
tached to the cars with a rope and then leaning backward while
pressing his feet against the railway ties. The cars together weighed
700 kN (about 80 tons). Assume that he pulled with a constant
force that was 2.5 times his body weight, at an upward angle uof
30from the horizontal. His mass was 80 kg, and he moved the cars
by 1.0 m. Neglecting any retarding force from the wheel rotation,
find the speed of the cars at the end of the pull.
WWWSSM
θ
Leg
joint Tibia
Rod
Figure 5-35 Problem 16.
m
Frictionless
θ
Figure 5-36 Problem 17.
to x(t)13.00 2.00t4.00t23.00t3, with xin meters and tin
seconds. In unit-vector notation, what is the net force acting on the
particle at t3.40 s?
••11 A 2.0 kg particle moves along an xaxis, being propelled by a
variable force directed along that axis. Its position is given by x
3.0 m (4.0 m/s)tct2(2.0 m/s3)t3, with xin meters and tin
seconds.The factor cis a constant.At t3.0 s, the force on the par-
ticle has a magnitude of 36 N and is in the negative direction of the
axis.What is c?
•••12 Two horizontal forces and act on a 4.0 kg disk that
slides over frictionless ice, on which an xy coordinate system is laid
out. Force is in the positive direction of the xaxis and has a mag-
nitude of 7.0 N. Force has a magnitude of 9.0 N. Figure 5-32
gives the xcomponent vxof the velocity of the disk as a function of
time tduring the sliding.What is the angle between the constant di-
rections of forces and ?F
:
2
F
:
1
F
:
2
F
:
1
F
:
2
F
:
1
•24 There are two horizontal
forces on the 2.0 kg box in the over-
head view of Fig. 5-38 but only one
(of magnitude F120 N) is shown.
The box moves along the xaxis. For
each of the following values for the acceleration axof the box,
find the second force in unit-vector notation: (a) 10 m/s2, (b) 20 m/s2,
(c) 0,(d) 10 m/s2,and (e) 20 m/s2.
•25 Sunjamming. A “sun yacht” is a spacecraft with a large sail
that is pushed by sunlight.Although such a push is tiny in everyday
circumstances, it can be large enough to send the spacecraft
outward from the Sun on a cost-free but slow trip. Suppose that
the spacecraft has a mass of 900 kg and receives a push of 20 N.
(a) What is the magnitude of the resulting acceleration? If the craft
starts from rest, (b) how far will it travel in 1 day and (c) how fast
will it then be moving?
•26 The tension at which a fishing line snaps is commonly called the
line’s “strength.”What minimum strength is needed for a line that is to
stop a salmon of weight 85 N in 11 cm if the fish is initially drifting at
2.8 m/s? Assume a constant deceleration.
•27 An electron with a speed of 1.2 107m/s moves hori-
zontally into a region where a constant vertical force of 4.5
1016 N acts on it. The mass of the electron is 9.11 1031 kg.
Determine the vertical distance the electron is deflected during the
time it has moved 30 mm horizontally.
•28 A car that weighs 1.30 104N is initially moving at
40 km/h when the brakes are applied and the car is brought to a
stop in 15 m. Assuming the force that stops the car is constant,
find (a) the magnitude of that force and (b) the time required for
the change in speed. If the initial speed is doubled, and the car ex-
periences the same force during the braking, by what factors are
(c) the stopping distance and (d) the stopping time multiplied?
(There could be a lesson here about the danger of driving at high
speeds.)
•29 A firefighter who weighs 712 N slides down a vertical pole
with an acceleration of 3.00 m/s2, directed downward.What are the
(a) magnitude and (b) direction (up or down) of the vertical force
on the firefighter from the pole and the (c) magnitude and (d) di-
rection of the vertical force on the pole from the firefighter?
•30 The high-speed winds around a tornado can drive pro-
jectiles into trees, building walls, and even metal traffic signs. In a
laboratory simulation, a standard wood toothpick was shot by
pneumatic gun into an oak branch.The toothpick’s mass was 0.13 g,
its speed before entering the branch was 220 m/s, and its penetra-
tion depth was 15 mm. If its speed was decreased at a uniform
rate, what was the magnitude of the force of the branch on the
toothpick?
••31 A block is projected up a frictionless inclined
plane with initial speed v03.50
m/s. The angle of incline is
32.0. (a) How far up the plane
does the block go? (b) How long
does it take to get there? (c) What is
its speed when it gets back to the
bottom?
••32 Figure 5-39 shows an overhead
view of a 0.0250 kg lemon half and
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118 CHAPTER 5 FORCE AND MOTION—I
x
F
1
Figure 5-38 Problem 24.
y
x
F1
F2
2
θ
1
θ
Figure 5-39 Problem 32.
•19 A 500 kg rocket sled can be accelerated at a constant
rate from rest to 1600 km/h in 1.8 s. What is the magnitude of the
required net force?
•20 A car traveling at 53 km/h hits a bridge abutment.A passen-
ger in the car moves forward a distance of 65 cm (with respect to
the road) while being brought to rest by an inflated air bag. What
magnitude of force (assumed constant) acts on the passenger’s up-
per torso, which has a mass of 41 kg?
•21 A constant horizontal force pushes a 2.00 kg FedEx pack-
age across a frictionless floor on which an xy coordinate system has
been drawn. Figure 5-37 gives the package’s xand yvelocity com-
ponents versus time t. What are the (a) magnitude and (b) direc-
tion of ?F
:
a
F
:
a
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•22 A customer sits in an amusement park ride in which the
compartment is to be pulled downward in the negative direction of
ayaxis with an acceleration magnitude of 1.24g, with g9.80 m/s2.
A 0.567 g coin rests on the customer’s knee. Once the motion be-
gins and in unit-vector notation, what is the coin’s acceleration rel-
ative to (a) the ground and (b) the customer? (c) How long does
the coin take to reach the compartment ceiling, 2.20 m above the
knee? In unit-vector notation, what are (d) the actual force on the
coin and (e) the apparent force according to the customer’s meas-
ure of the coin’s acceleration?
•23 Tarzan, who weighs 820 N, swings from a cliff at the end of a
20.0 m vine that hangs from a high tree limb and initially makes an
angle of 22.0with the vertical.Assume that an xaxis extends hori-
zontally away from the cliff edge and a yaxis extends upward.
Immediately after Tarzan steps off the cliff, the tension in the vine
is 760 N. Just then, what are (a) the force on him from the vine in
unit-vector notation and the net force on him (b) in unit-vector no-
tation and as (c) a magnitude and (d) an angle relative to the
positive direction of the xaxis? What are the (e) magnitude and
(f) angle of Tarzan’s acceleration just then?
vx (m/s)
vy (m/s)
t (s)
3210
5
10
t (s)
3210
–5
–10
0
Figure 5-37 Problem 21.
stant speed up a frictionless ramp
( 30.0) by a horizontal force
. What are the magnitudes of (a)
and (b) the force on the crate from
the ramp?
••35 The velocity of a 3.00 kg parti-
cle is given by (8.00t+ 3.00t2)
m/s, with time tin seconds.At the instant the net force on the parti-
cle has a magnitude of 35.0 N, what are the direction (relative to
the positive direction of the xaxis) of (a) the net force and (b) the
particle’s direction of travel?
••36 Holding on to a towrope moving parallel to a frictionless ski
slope, a 50 kg skier is pulled up the slope, which is at an angle of
8.0with the horizontal.What is the magnitude Frope of the force on
the skier from the rope when (a) the magnitude vof the skier’s ve-
locity is constant at 2.0 m/s and (b) v2.0 m/s as vincreases at a
rate of 0.10 m/s2?
••37 A 40 kg girl and an 8.4 kg sled are on the frictionless ice of a
frozen lake, 15 m apart but connected by a rope of negligible mass.
The girl exerts a horizontal 5.2 N force on the rope. What are the ac-
celeration magnitudes of (a) the sled and (b) the girl? (c) How far
from the girl’s initial position do they meet?
••38 A 40 kg skier skis directly down a frictionless slope angled
at 10to the horizontal.Assume the skier moves in the negative di-
rection of an xaxis along the slope. A wind force with component
Fxacts on the skier.What is Fxif the magnitude of the skier’s veloc-
ity is (a) constant, (b) increasing at a rate of 1.0 m/s2, and (c) in-
creasing at a rate of 2.0 m/s2?
••39 A sphere of mass 3.0 104kg is suspended from
a cord. A steady horizontal breeze pushes the sphere so that the
cord makes a constant angle of 37with the vertical. Find (a) the
push magnitude and (b) the tension in the cord.
••40 A dated box of dates, of mass 5.00 kg, is sent sliding up a
frictionless ramp at an angle of to the horizontal. Figure 5-41 gives,
ILW
j
ˆ
i
ˆ
v
:
F
:
F
:
m100 kg is pushed at con-
119
PROBLEMS
has the constant velocity and (c) has the
varying velocity where tis time?
••33 An elevator cab and its load have a combined mass of 1600 kg.
Find the tension in the supporting cable when the cab, originally
moving downward at 12 m/s, is brought to rest with constant accel-
eration in a distance of 42 m.
••34 In Fig. 5-40, a crate of mass
(13.0ti
ˆ14.0tj
ˆ) m/s2,v
:
v
:(13.0i
ˆ14.0j
ˆ) m/s,
has a magnitude of 7.00 N and is at 30.0. In unit-vector no-
tation, what is the third force if the lemon half (a) is stationary, (b)
2F
:
2
two of the three horizontal forces that act on it as it is on a frictionless
table. Force has a magnitude of 6.00 N and is at 30.0. Force
1F
:
1
m
θ
F
Figure 5-40 Problem 34.
vx (m/s)
t (s)
4
2
–2
–4
1023
Figure 5-41 Problem 40.
as a function of time t,the component vxof the box’s velocity along an
xaxis that extends directly up the ramp.What is the magnitude of the
normal force on the box from the ramp?
••41 Using a rope that will snap if the tension in it exceeds 387 N,
you need to lower a bundle of old roofing material weighing 449 N
from a point 6.1 m above the ground. Obviously if you hang the bun-
dle on the rope, it will snap. So, you allow the bundle to accelerate
downward. (a) What magnitude of the bundle’s acceleration will put
the rope on the verge of snapping? (b) At that acceleration, with
what speed would the bundle hit the ground?
••42 In earlier days, horses pulled barges down canals in the
manner shown in Fig. 5-42. Suppose the horse pulls on the rope
with a force of 7900 N at an angle of u18to the direction of
motion of the barge, which is headed straight along the positive
direction of an xaxis. The mass of the barge is 9500 kg, and the
magnitude of its acceleration is 0.12 m/s2.What are the (a) magni-
tude and (b) direction (relative to positive x) of the force on the
barge from the water?
θ
Figure 5-42 Problem 42.
••43 In Fig. 5-43, a chain consisting of five
links, each of mass 0.100 kg, is lifted vertically
with constant acceleration of magnitude a2.50
m/s2. Find the magnitudes of (a) the force on link
1 from link 2, (b) the force on link 2 from link 3,
(c) the force on link 3 from link 4, and (d) the
force on link 4 from link 5. Then find the magni-
tudes of (e) the force on the top link from the
person lifting the chain and (f) the net force accel-
erating each link.
••44 A lamp hangs vertically from a cord in a de-
scending elevator that decelerates at 2.4 m/s2. (a)
If the tension in the cord is 89 N, what is the lamp’s
mass? (b) What is the cord’s tension when the ele-
vator ascends with an upward acceleration of 2.4 m/s2?
••45 An elevator cab that weighs 27.8 kN moves upward.What is
the tension in the cable if the cab’s speed is (a) increasing at a rate
of 1.22 m/s2and (b) decreasing at a rate of 1.22 m/s2?
••46 An elevator cab is pulled upward by a cable.The cab and its
single occupant have a combined mass of 2000 kg.When that occu-
pant drops a coin, its acceleration relative to the cab is 8.00 m/s2
downward.What is the tension in the cable?
••47 The Zacchini family was renowned for their hu-
man-cannonball act in which a family member was shot from a
cannon using either elastic bands or compressed air. In one version
of the act, Emanuel Zacchini was shot over three Ferris wheels to
land in a net at the same height as the open end of the cannon and
at a range of 69 m. He was propelled inside the barrel for 5.2 m and
launched at an angle of 53. If his mass was 85 kg and he underwent
constant acceleration inside the barrel, what was the magnitude of
the force propelling him? (Hint: Treat the launch as though it were
along a ramp at 53. Neglect air drag.)
F
:
SSM
F
1
2
3
4
5
a
Figure 5-43
Problem 43.
120 CHAPTER 5 FORCE AND MOTION—I
B
A
F
a
(
a
)
B
A
F
a
(
b
)
Figure 5-51 Problem 56.
••48 In Fig. 5-44, elevator cabs Aand Bare con-
nected by a short cable and can be pulled upward or
lowered by the cable above cab A. Cab Ahas mass
1700 kg; cab Bhas mass 1300 kg.A 12.0 kg box of cat-
nip lies on the floor of cab A.The tension in the cable
connecting the cabs is 1.91 104N.What is the mag-
nitude of the normal force on the box from the floor?
••49 In Fig. 5-45, a block of mass m5.00 kg is
pulled along a horizontal frictionless floor by a cord
that exerts a force of magnitude F12.0 N at an
angle u25.0. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude Fis
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before it is lifted
(completely) off the floor?
••54 Figure 5-49 shows four penguins that are being playfully
pulled along very slippery (frictionless) ice by a curator.The masses
of three penguins and the tension in two of the cords are m112 kg,
m315 kg, m420 kg, T2111 N, and T4222 N. Find the pen-
guin mass m2that is not given.
••50 In Fig. 5-46, three ballot
boxes are connected by cords, one
of which wraps over a pulley having
negligible friction on its axle and
negligible mass. The three masses
are mA30.0 kg, mB40.0 kg,
and mC10.0 kg.When the assem-
bly is released from rest, (a) what is the tension in the cord con-
necting Band C, and (b) how far does Amove in the first 0.250 s
(assuming it does not reach the pulley)?
••51 Figure 5-47 shows two blocks connected by
a cord (of negligible mass) that passes over a fric-
tionless pulley (also of negligible mass). The
arrangement is known as Atwood’s machine. One
block has mass m11.30 kg; the other has mass m2
2.80 kg.What are (a) the magnitude of the blocks’ ac-
celeration and (b) the tension in the cord?
••52 An 85 kg man lowers himself to the ground
from a height of 10.0 m by holding onto a rope that
runs over a frictionless pulley to a 65 kg sandbag.
With what speed does the man hit the ground if he
started from rest?
••53 In Fig. 5-48, three connected blocks are
pulled to the right on a horizontal frictionless table
by a force of magnitude T3 65.0 N. If m1 12.0 kg,
m2 24.0 kg, and m3 31.0 kg, calculate (a) the magnitude of the
system’s acceleration, (b) the tension T1, and (c) the tension T2.
A
B
Figure 5-44
Problem 48.
F
θ
m
Figure 5-45
Problems 49 and 60.
A
B
C
Figure 5-46 Problem 50.
m1m2m3
T1T2T3
Figure 5-48 Problem 53.
m1
m2
Figure 5-47
Problems 51
and 65.
m1m3
m4
T2T4
Figure 5-49 Problem 54.
••55 Two blocks are in
contact on a frictionless table. A horizon-
tal force is applied to the larger block, as
shown in Fig. 5-50. (a) If m12.3 kg,
m21.2 kg, and F3.2 N, find the mag-
nitude of the force between the two
blocks. (b) Show that if a force of the same
magnitude Fis applied to the smaller
block but in the opposite direction, the magnitude of the force be-
tween the blocks is 2.1 N, which is not the same value calculated in
(a). (c) Explain the difference.
••56 In Fig. 5-51a, a constant horizontal force is applied to
block A, which pushes against block Bwith a 20.0 N force directed
horizontally to the right. In Fig. 5-51b, the same force is applied
to block B; now block Apushes on block Bwith a 10.0 N force
directed horizontally to the left. The blocks have a combined mass
of 12.0 kg. What are the magnitudes of (a) their acceleration in
Fig. 5-51aand (b) force ?F
:
a
F
:
a
F
:
a
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m1
m2
F
Figure 5-50
Problem 55.
••57 A block of mass m13.70 kg on a frictionless plane in-
ILW
clined at angle 30.0is connected by a cord over a massless,
frictionless pulley to a second block of mass m22.30 kg (Fig.
5-52). What are (a) the magnitude of the acceleration of each
block, (b) the direction of the acceleration of the hanging block,
and (c) the tension in the cord?
m2
θ
m1
Figure 5-52 Problem 57.
••58 Figure 5-53 shows a man sitting in a bosun’s chair that dan-
gles from a massless rope, which runs over a massless, frictionless
pulley and back down to the man’s hand. The combined mass of
man and chair is 95.0 kg.With what force magnitude must the man
pull on the rope if he is to rise (a) with a constant velocity and
121
PROBLEMS
(b) with an upward acceleration of
1.30 m/s2? (Hint: A free-body dia-
gram can really help.) If the rope
on the right extends to the ground
and is pulled by a co-worker, with
what force magnitude must the co-
worker pull for the man to rise (c)
with a constant velocity and (d)
with an upward acceleration of
1.30 m/s2? What is the magnitude
of the force on the ceiling from the
pulley system in (e) part a, (f) part
b, (g) part c, and (h) part d?
••59 A 10 kg monkey climbs
up a massless rope that runs over a
frictionless tree limb and back
down to a 15 kg package on the
ground (Fig. 5-54). (a) What is the
magnitude of the least acceleration
the monkey must have if it is to lift
the package off the ground? If, after
the package has been lifted, the
monkey stops its climb and holds
onto the rope, what are the (b)
magnitude and (c) direction of the
monkey’s acceleration and (d) the
tension in the rope?
••60 Figure 5-45 shows a 5.00 kg
block being pulled along a friction-
less floor by a cord that applies a
force of constant magnitude 20.0 N
but with an angle u(t) that varies
with time.When angle u25.0,at
what rate is the acceleration of the
block changing if (a) u(t)
(2.00 102deg/s)tand (b) u(t)(2.00 102deg/s)t? (Hint:
The angle should be in radians.)
••61 A hot-air balloon of mass Mis descending vertically
with downward acceleration of magnitude a.How much mass (ballast)
must be thrown out to give the balloon an upward acceleration of mag-
nitude a? Assume that the upward force from the air (the lift) does not
change because of the decrease in mass.
•••62 In shot putting, many athletes elect to launch the shot
ILWSSM
SSM
the axis, with a speed of 3.0 m/s.What are its (a) speed and (b) direc-
tion of travel at t11 s?
Figure 5-53 Problem 58.
Bananas
Figure 5-54 Problem 59.
6
0
2 4 6 8 10 12
–4
t (s)
F
x (N)
Figure 5-55 Problem 63.
at an angle that is smaller than the theoretical one (about 42) at
which the distance of a projected ball at the same speed and
height is greatest. One reason has to do with the speed the athlete
can give the shot during the acceleration phase of the throw.
Assume that a 7.260 kg shot is accelerated along a straight path of
length 1.650 m by a constant applied force of magnitude 380.0 N,
starting with an initial speed of 2.500 m/s (due to the athlete’s pre-
liminary motion). What is the shot’s speed at the end of the accel-
eration phase if the angle between the path and the horizontal is
(a) 30.00and (b) 42.00? (Hint: Treat the motion as though it
were along a ramp at the given angle.) (c) By what percent is the
launch speed decreased if the athlete increases the angle from
30.00to 42.00?
•••63 Figure 5-55 gives, as a function of time t, the force compo-
nent Fxthat acts on a 3.00 kg ice block that can move only along
the xaxis. At t0, the block is moving in the positive direction of
•••64 Figure 5-56 shows a box of mass m21.0 kg on a fric-
θ
F
m
2
m
1
Figure 5-56 Problem 64.
•••65 Figure 5-47 shows Atwood’s machine, in which two con-
tainers are connected by a cord (of negligible mass) passing over a
frictionless pulley (also of negligible mass).At time t0, container
1 has mass 1.30 kg and container 2 has mass 2.80 kg, but container 1
is losing mass (through a leak) at the constant rate of 0.200 kg/s. At
what rate is the acceleration magnitude of the containers changing
at (a) t0 and (b) t3.00 s? (c) When does the acceleration reach
its maximum value?
•••66 Figure 5-57 shows a section of a cable-car system. The
maximum permissible mass of each car with occupants is 2800 kg.
The cars, riding on a support cable, are pulled by a second cable
attached to the support tower on each car.Assume that the cables
Support cable
Pull cable
θ
Figure 5-57 Problem 66.
tionless plane inclined at angle u30. It is connected by a cord of
negligible mass to a box of mass m13.0 kg on a horizontal fric-
tionless surface. The pulley is frictionless and massless. (a) If the
magnitude of horizontal force is 2.3 N, what is the tension in the
connecting cord? (b) What is the largest value the magnitude of
may have without the cord becoming slack?
F
:
F
:
antioxidants (m11.0 kg) on a fric-
tionless inclined surface is con-
nected to a tin of corned beef (m2
2.0 kg). The pulley is massless and
frictionless. An upward force of
magnitude F6.0 N acts on the
corned beef tin, which has a down-
ward acceleration of 5.5 m/s2. What
are (a) the tension in the connecting
cord and (b) angle b?
74 The only two forces acting on a
body have magnitudes of 20 N and
35
N and directions that differ by
80. The resulting acceleration has a
magnitude of 20 m/s2. What is the
mass of the body?
75 Figure 5-62 is an overhead
view of a 12 kg tire that is to be
pulled by three horizontal ropes.
One rope’s force (F150 N) is in-
dicated. The forces from the other
ropes are to be oriented such that
the tire’s acceleration magnitude ais
least. What is that least aif (a) F2
30 N, F320 N; (b) F230 N, F3
10 N;and (c) F2F330 N?
76 A block of mass Mis pulled
along a horizontal frictionless sur-
face by a rope of mass m, as shown
in Fig. 5-63. A horizontal force
acts on one end of the rope.
(a) Show that the rope must sag, even if only by an imperceptible
amount. Then, assuming that the sag is negligible, find (b) the ac-
celeration of rope and block, (c) the force on the block from the
rope, and (d) the tension in the rope at its midpoint.
77 A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force of magni-
tude F450 N on the rope, which is inclined at an upward angle
u38to the horizontal, and the floor exerts a horizontal force
of magnitude f125 N that opposes the motion. Calculate the
magnitude of the acceleration of the crate if (a) its mass is 310 kg
and (b) its weight is 310 N.
78 In Fig. 5-64, a force of mag-
nitude 12 N is applied to a FedEx
box of mass m21.0 kg. The force
is directed up a plane tilted by u
37. The box is connected by a cord
to a UPS box of mass m13.0 kg
on the floor. The floor, plane, and
pulley are frictionless, and the
masses of the pulley and cord are negligible. What is the tension in
the cord?
79 A certain particle has a weight of 22 N at a point where
g9.8 m/s2. What are its (a) weight and (b) mass at a point where
g4.9 m/s2? What are its (c) weight and (d) mass if it is moved to
a point in space where g0?
80 An 80 kg person is parachuting and experiencing a downward
acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg. (a)
F
:
SSM
F
:
122 CHAPTER 5 FORCE AND MOTION—I
F
m
M
Figure 5-63 Problem 76.
θ
F
m2
m1
Figure 5-64 Problem 78.
are taut and inclined at angle u35. What is the difference in
tension between adjacent sections of pull cable if the cars are at
the maximum permissible mass and are being accelerated up the
incline at 0.81 m/s2?
•••67 Figure 5-58 shows three
blocks attached by cords that loop
over frictionless pulleys. Block B
lies on a frictionless table; the
masses are mA6.00 kg, mB8.00
kg, and mC10.0 kg. When the
blocks are released, what is the
tension in the cord at the right?
•••68 A shot putter launches a 7.260 kg shot by pushing it
along a straight line of length 1.650 m and at an angle of 34.10°
from the horizontal, accelerating the shot to the launch speed
from its initial speed of 2.500 m/s (which is due to the athlete’s
preliminary motion).The shot leaves the hand at a height of 2.110 m
and at an angle of 34.10, and it lands at a horizontal distance of
15.90 m. What is the magnitude of the athlete’s average force on
the shot during the acceleration phase? (Hint: Treat the motion
during the acceleration phase as though it were along a ramp at
the given angle.)
Additional Problems
69 In Fig. 5-59, 4.0 kg block Aand 6.0 kg block Bare connected by
a string of negligible mass. Force acts on block A;
force acts on block B.What is the tension in the string?F
:
B(24 N)i
ˆF
:
A(12 N)i
ˆ
73 In Fig. 5-61, a tin of
SSM
70 An 80 kg man drops to a concrete patio from a window
0.50 m above the patio.He neglects to bend his knees on landing, tak-
ing 2.0 cm to stop. (a) What is his average acceleration from when his
feet first touch the patio to when he stops? (b) What is the magnitude
of the average stopping force exerted on him by the patio?
71 Figure 5-60 shows a box of dirty money (mass m13.0 kg)
SSM
A
B
C
Figure 5-58 Problem 67.
x
A B
FAFB
Figure 5-59 Problem 69.
1
θ
2
θ
m
2
m
1
Figure 5-60 Problem 71.
β
F
m1
m2
Figure 5-61 Problem 73.
72 Three forces act on a particle that moves with unchanging ve-
locity Two of the forces are
and .What is
the third force?
(8 N)j
ˆ(2 N)k
ˆ
F2
:(5 N)i
ˆ(3 N)j
ˆ(2 N)k
ˆF1
:(2 N)i
ˆv
:(2 m/s)i
ˆ(7 m/s)j
ˆ.
x
F1
Figure 5-62 Problem 75.
on a frictionless plane inclined at angle 30. The box is con-
nected via a cord of negligible mass to a box of laundered money
(mass m22.0 kg) on a frictionless plane inclined at angle u260.
The pulley is frictionless and has negligible mass. What is the ten-
sion in the cord?
1
123
PROBLEMS
What is the upward force on the open parachute from the air? (b)
What is the downward force on the parachute from the person?
81 A spaceship lifts off vertically from the Moon, where g
1.6 m/s2. If the ship has an upward acceleration of 1.0 m/s2as it lifts
off, what is the magnitude of the force exerted by the ship on its pi-
lot, who weighs 735 N on Earth?
82 In the overhead view of Fig.
5-65, five forces pull on a box of
mass m4.0 kg. The force magni-
tudes are F111 N, F217 N,
F33.0 N, F414 N, and F55.0 N,
and angle u4is 30. Find the box’s
acceleration (a) in unit-vector nota-
tion and as (b) a magnitude and
(c) an angle relative to the positive
direction of the xaxis.
83 A certain force gives an
object of mass m1an acceleration
of 12.0 m/s2and an object of mass m2an acceleration of 3.30
m/s2. What acceleration would the force give to an object of mass
(a) m2m1and (b) m2m1?
84 You pull a short refrigerator with a constant force across a
greased (frictionless) floor, either with horizontal (case 1) or with
tilted upward at an angle u(case 2). (a) What is the ratio of the re-
frigerator’s speed in case 2 to its speed in case 1 if you pull for a cer-
tain time t? (b) What is this ratio if you pull for a certain distance d?
85 A 52 kg circus performer is to slide down a rope that will
break if the tension exceeds 425 N. (a) What happens if the per-
former hangs stationary on the rope? (b) At what magnitude of ac-
celeration does the performer just avoid breaking the rope?
86 Compute the weight of a 75 kg space ranger (a) on Earth,
(b) on Mars, where g3.7 m/s2, and (c) in interplanetary space,
where g0. (d) What is the ranger’s mass at each location?
87 An object is hung from a spring balance attached to the ceil-
ing of an elevator cab. The balance reads 65 N when the cab is
standing still. What is the reading when the cab is moving upward
(a) with a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s
while decelerating at a rate of 2.4 m/s2?
88 Imagine a landing craft approaching the surface of Callisto,
one of Jupiter’s moons. If the engine provides an upward force
(thrust) of 3260 N, the craft descends at constant speed; if the en-
gine provides only 2200 N, the craft accelerates downward at
0.39 m/s2. (a) What is the weight of the landing craft in the vicinity
of Callisto’s surface? (b) What is the mass of the craft? (c) What is
the magnitude of the free-fall acceleration near the surface of
Callisto?
89 A 1400 kg jet engine is fastened to the fuselage of a passenger
jet by just three bolts (this is the usual practice). Assume that each
bolt supports one-third of the load. (a) Calculate the force on each
bolt as the plane waits in line for clearance to take off. (b) During
flight, the plane encounters turbulence, which suddenly imparts an
upward vertical acceleration of 2.6 m/s2to the plane. Calculate the
force on each bolt now.
90 An interstellar ship has a mass of 1.20 106kg and is initially at
rest relative to a star system. (a) What constant acceleration is needed
to bring the ship up to a speed of 0.10c(where cis the speed of light,
3.0 108m/s) relative to the star system in 3.0 days? (b) What is that
F
:F
:F
:
SSM
acceleration in gunits? (c) What force is required for the accelera-
tion? (d) If the engines are shut down when 0.10cis reached (the
speed then remains constant), how long does the ship take (start to
finish) to journey 5.0 light-months, the distance that light travels in
5.0 months?
91 A motorcycle and 60.0 kg rider accelerate at 3.0 m/s2up
SSM
x
y
F1
F2
F4
F5
F3
4
θ
Figure 5-65 Problem 82.
94 For sport, a 12 kg armadillo runs onto a large pond of level,
frictionless ice. The armadillo’s initial velocity is 5.0 m/s along the
positive direction of an xaxis. Take its initial position on the ice as
being the origin. It slips over the ice while being pushed by a wind
with a force of 17 N in the positive direction of the yaxis. In unit-
vector notation, what are the animal’s (a) velocity and (b) position
vector when it has slid for 3.0 s?
95 Suppose that in Fig. 5-12, the masses of the blocks are 2.0 kg
and 4.0 kg. (a) Which mass should the hanging block have if the
magnitude of the acceleration is to be as large as possible? What
then are (b) the magnitude of the acceleration and (c) the tension
in the cord?
96 A nucleus that captures a stray neutron must bring the neu-
tron to a stop within the diameter of the nucleus by means of the
strong force. That force, which “glues” the nucleus together, is ap-
proximately zero outside the nucleus. Suppose that a stray neutron
with an initial speed of 1.4 107m/s is just barely captured by a
nucleus with diameter d1.0 1014 m. Assuming the strong
force on the neutron is constant, find the magnitude of that force.
The neutron’s mass is 1.67 1027 kg.
97 If the 1 kg standard body is accelerated by only
, then what
is (a) in unit-vector notation and as (b) a magnitude and
(c) an angle relative to the positive xdirection? What are the (d)
magnitude and (e) angle of ?a
:
F
:
net
(3.0 N)i
ˆ(4.0 N)j
ˆ and F
:
2(2.0 N)i
ˆ(6.0 N)j
ˆF
:
1
(a) (b)
m1
m2
m3
m5
Figure 5-66 Problem 93.
a ramp inclined 10above the horizontal.What are the magnitudes
of (a) the net force on the rider and (b) the force on the rider from
the motorcycle?
92 Compute the initial upward acceleration of a rocket of mass
1.3 104kg if the initial upward force produced by its engine (the
thrust) is 2.6 105N. Do not neglect the gravitational force on the
rocket.
93 Figure 5-66ashows a mobile hanging from a ceiling; it
consists of two metal pieces (m13.5 kg and m24.5 kg) that are
strung together by cords of negligible mass. What is the tension in
(a) the bottom cord and (b) the top cord? Figure 5-66bshows a
mobile consisting of three metal pieces.Two of the masses are m3
4.8 kg and m55.5 kg.The tension in the top cord is 199 N.What is
the tension in (c) the lowest cord and (d) the middle cord?
SSM
124
What Is Physics?
In this chapter we focus on the physics of three common types of force: frictional
force, drag force, and centripetal force. An engineer preparing a car for the
Indianapolis 500 must consider all three types. Frictional forces acting on the tires
are crucial to the car’s acceleration out of the pit and out of a curve (if the car hits
an oil slick, the friction is lost and so is the car). Drag forces acting on the car
from the passing air must be minimized or else the car will consume too much
fuel and have to pit too early (even one 14 s pit stop can cost a driver the race).
Centripetal forces are crucial in the turns (if there is insufficient centripetal force,
the car slides into the wall).We start our discussion with frictional forces.
Friction
Frictional forces are unavoidable in our daily lives. If we were not able to coun-
teract them, they would stop every moving object and bring to a halt every
rotating shaft. About 20% of the gasoline used in an automobile is needed to
counteract friction in the engine and in the drive train. On the other hand, if fric-
tion were totally absent, we could not get an automobile to go anywhere, and we
could not walk or ride a bicycle. We could not hold a pencil, and, if we could, it
would not write. Nails and screws would be useless, woven cloth would fall apart,
and knots would untie.
CHAPTER 6
Force and Motion—II
6-1FRICTION
After reading this module, you should be able to . . .
6.01 Distinguish between friction in a static situation and a
kinetic situation.
6.02 Determine direction and magnitude of a frictional force.
6.03 For objects on horizontal, vertical, or inclined planes in
situations involving friction, draw free-body diagrams and
apply Newton’s second law.
●When a force tends to slide a body along a surface, a fric-
tional force from the surface acts on the body. The frictional
force is parallel to the surface and directed so as to oppose the
sliding. It is due to bonding between the body and the surface.
If the body does not slide, the frictional force is a static
frictional force . If there is sliding, the frictional force is a
kinetic frictional force .
●If a body does not move, the static frictional force and
the component of parallel to the surface are equal in magni-
tude, and is directed opposite that component. If the com-
ponent increases, fsalso increases.
f
:
s
F
:f
:
s
f
:
k
f
:
s
F
:
●The magnitude of has a maximum value s,max given by
fs,max msFN,
where msis the coefficient of static friction and FNis the mag-
nitude of the normal force. If the component of parallel to
the surface exceeds fs,max, the body slides on the surface.
●If the body begins to slide on the surface, the magnitude of the
frictional force rapidly decreases to a constant value given by
fkmkFN,
where mkis the coefficient of kinetic friction.
fk
F
:
f
:
f
:
s
Key Ideas
Learning Objectives
125
6-1 FRICTION
(a)
(b)
(c)
(d)
fs
fs
fs
Fg
Fg
Fg
Fg
F
F
F
FN
FN
FN
FN
There is no attempt
at sliding. Thus,
no friction and
no motion.
Frictional force = 0
Force F attempts
sliding but is balanced
by the frictional force.
No motion.
Force F is now
stronger but is still
balanced by the
frictional force.
No motion.
Force F is now even
stronger but is still
balanced by the
frictional force.
No motion.
Frictional force = F
Frictional force = F
Frictional force = F
Figure 6-1 (a) The forces on a
stationary block. (b–d) An external
force , applied to the block, is
balanced by a static frictional force
. As Fis increased, fsalso increases,
until fsreaches a certain maximum
value. (Figure continues)
f
:
s
F
:
A
Three Experiments. Here we deal with the frictional forces that exist be-
tween dry solid surfaces, either stationary relative to each other or moving across
each other at slow speeds. Consider three simple thought experiments:
1. Send a book sliding across a long horizontal counter. As expected, the book
slows and then stops. This means the book must have an acceleration parallel
to the counter surface, in the direction opposite the book’s velocity. From
Newton’s second law, then, a force must act on the book parallel to the counter
surface, in the direction opposite its velocity.That force is a frictional force.
2. Push horizontally on the book to make it travel at constant velocity along the
counter. Can the force from you be the only horizontal force on the book?
No, because then the book would accelerate. From Newton’s second law, there
must be a second force, directed opposite your force but with the same magni-
tude, so that the two forces balance. That second force is a frictional force,
directed parallel to the counter.
3. Push horizontally on a heavy crate. The crate does not move. From Newton’s
second law, a second force must also be acting on the crate to counteract your
force. Moreover, this second force must be directed opposite your force and
have the same magnitude as your force, so that the two forces balance. That
second force is a frictional force. Push even harder. The crate still does not
move.Apparently the frictional force can change in magnitude so that the two
forces still balance. Now push with all your strength.The crate begins to slide.
Evidently, there is a maximum magnitude of the frictional force. When you
exceed that maximum magnitude, the crate slides.
Two Types of Friction. Figure 6-1 shows a similar situation. In Fig. 6-1a, a block
rests on a tabletop, with the gravitational force balanced by a normal force F
:
N.F
:
g
In Fig. 6-1b, you exert a force on the block, attempting to pull it to the left. In re-
sponse, a frictional force is directed to the right, exactly balancing your force.
The force is called the static frictional force. The block does not move.f
:
s
f
:
s
F
:
126 CHAPTER 6 FORCE AND MOTION—II
Figures 6-1cand 6-1dshow that as you increase the magnitude of your
applied force, the magnitude of the static frictional force also increases and
the block remains at rest. When the applied force reaches a certain magnitude,
however, the block “breaks away” from its intimate contact with the tabletop and
accelerates leftward (Fig. 6-1e).The frictional force that then opposes the motion
is called the kinetic frictional force .
Usually, the magnitude of the kinetic frictional force, which acts when there is
motion, is less than the maximum magnitude of the static frictional force, which
acts when there is no motion. Thus, if you wish the block to move across the sur-
face with a constant speed, you must usually decrease the magnitude of the
applied force once the block begins to move, as in Fig. 6-1f. As an example,
Fig. 6-1gshows the results of an experiment in which the force on a block was
slowly increased until breakaway occurred. Note the reduced force needed to
keep the block moving at constant speed after breakaway.
Microscopic View. A frictional force is, in essence, the vector sum of many
forces acting between the surface atoms of one body and those of another body. If
two highly polished and carefully cleaned metal surfaces are brought together in
a very good vacuum (to keep them clean), they cannot be made to slide over each
other. Because the surfaces are so smooth, many atoms of one surface contact
many atoms of the other surface, and the surfaces cold-weld together instantly,
forming a single piece of metal. If a machinist’s specially polished gage blocks are
brought together in air, there is less atom-to-atom contact, but the blocks stick
firmly to each other and can be separated only by means of a wrenching motion.
Usually, however, this much atom-to-atom contact is not possible. Even a highly
polished metal surface is far from being flat on the atomic scale. Moreover, the
surfaces of everyday objects have layers of oxides and other contaminants that
reduce cold-welding.
When two ordinary surfaces are placed together, only the high points touch
each other.(It is like having the Alps of Switzerland turned over and placed down
on the Alps of Austria.) The actual microscopic area of contact is much less than
the apparent macroscopic contact area, perhaps by a factor of 104. Nonetheless,
f
:
k
f
:
s
(e)
(f)
fk
fk
F
Time
Magnitude of
frictional force
Maximum value of fs
fkis approximately
constant
Breakaway
(g)
0
Fg
F
Fg
a
v
FN
FN
Finally, the applied force
has overwhelmed the
static frictional force.
Block slides and
accelerates.
Static frictional force
can only match growing
applied force.
Weak kinetic
frictional force
Same weak kinetic
frictional force
Kinetic frictional force
has only one value
(no matching).
To maintain the speed,
weaken force F to match
the weak frictional force.
Figure 6-1 (Continued) (e) Once fsreaches
its maximum value, the block “breaks
away,” accelerating suddenly in the direc-
tion of . (f) If the block is now to move
with constant velocity, Fmust be reduced
from the maximum value it had just
before the block broke away. (g) Some
experimental results for the sequence
(a) through (f). In WileyPLUS, this
figure is available as an animation with
voiceover.
F
:
many contact points do cold-weld together. These welds produce static friction
when an applied force attempts to slide the surfaces relative to each other.
If the applied force is great enough to pull one surface across the other, there
is first a tearing of welds (at breakaway) and then a continuous re-forming and
tearing of welds as movement occurs and chance contacts are made (Fig. 6-2).
The kinetic frictional force that opposes the motion is the vector sum of the
forces at those many chance contacts.
If the two surfaces are pressed together harder, many more points cold-weld.
Now getting the surfaces to slide relative to each other requires a greater applied
force: The static frictional force has a greater maximum value. Once the sur-
faces are sliding, there are many more points of momentary cold-welding, so the
kinetic frictional force also has a greater magnitude.
Often, the sliding motion of one surface over another is “jerky” because the two
surfaces alternately stick together and then slip. Such repetitive stick-and-slip can pro-
duce squeaking or squealing, as when tires skid on dry pavement, fingernails scratch
along a chalkboard, or a rusty hinge is opened. It can also produce beautiful and capti-
vating sounds, as in music when a bow is drawn properly across a violin string.
Properties of Friction
Experiment shows that when a dry and unlubricated body presses against a surface
in the same condition and a force attempts to slide the body along the surface,
the resulting frictional force has three properties:
Property 1. If the body does not move, then the static frictional force and the
component of that is parallel to the surface balance each other. They are
equal in magnitude, and is directed opposite that component of .
Property 2. The magnitude of has a maximum value fs,max that is given by
fs,max msFN, (6-1)
where msis the coefficient of static friction and FNis the magnitude of the
normal force on the body from the surface. If the magnitude of the compo-
nent of that is parallel to the surface exceeds fs,max, then the body begins to
slide along the surface.
Property 3. If the body begins to slide along the surface, the magnitude of the
frictional force rapidly decreases to a value fkgiven by
fkmkFN, (6-2)
where mkis the coefficient of kinetic friction. Thereafter, during the sliding, a
kinetic frictional force with magnitude given by Eq. 6-2 opposes the motion.
The magnitude FNof the normal force appears in properties 2 and 3 as a
measure of how firmly the body presses against the surface. If the body presses
harder, then, by Newton’s third law, FNis greater. Properties 1 and 2 are worded
in terms of a single applied force , but they also hold for the net force of several
applied forces acting on the body. Equations 6-1 and 6-2 are not vector equations;
the direction of or is always parallel to the surface and opposed to the at-
tempted sliding,and the normal force is perpendicular to the surface.
The coefficients msand mkare dimensionless and must be determined experi-
mentally. Their values depend on certain properties of both the body and the
surface; hence, they are usually referred to with the preposition “between,” as in
“the value of msbetween an egg and a Teflon-coated skillet is 0.04, but that between
rock-climbing shoes and rock is as much as 1.2.” We assume that the value of mk
does not depend on the speed at which the body slides along the surface.
F
:
N
f
:
k
f
:
s
F
:
f
:
k
F
:
f
:
s
F
:
f
:
s
F
:f
:
s
F
:
f
:
k
f
:
s
f
:
k
127
6-1 FRICTION
Figure 6-2 The mechanism of sliding
friction. (a) The upper surface is sliding to
the right over the lower surface in this
enlarged view. (b) A detail, showing two
spots where cold-welding has occurred.
Force is required to break the welds and
maintain the motion.
(a)
(b)
128 CHAPTER 6 FORCE AND MOTION—II
Checkpoint 1
A block lies on a floor.(a) What is the magnitude of the frictional force on it from the
floor? (b) If a horizontal force of 5 N is now applied to the block, but the block does
not move, what is the magnitude of the frictional force on it? (c) If the maximum
value fs,max of the static frictional force on the block is 10 N, will the block move if the
magnitude of the horizontally applied force is 8 N? (d) If it is 12 N? (e) What is the
magnitude of the frictional force in part (c)?
ond law as
FNmg Fsin um(0), (6-4)
which gives us
FNmg Fsin u. (6-5)
Now we can evaluate fs,max msFN:
fs,max ms(mg Fsin u)
(0.700)((8.00 kg)(9.8 m/s2)(12.0 N)(sin 30))
59.08 N. (6-6)
Because the magnitude Fx(10.39 N) of the force com-
ponent attempting to slide the block is less than fs,max
(59.08 N), the block remains stationary.That means that
the magnitude fsof the frictional force matches Fx. From
Fig. 6-3d, we can write Newton’s second law for xcompo-
nents as
Fx fsm(0), (6-7)
and thus fs Fx10.39 N 10.4 N. (Answer)
Sample Problem 6.01 Angled force applied to an initially stationary block
This sample problem involves a tilted applied force,
which requires that we work with components to find a
frictional force. The main challenge is to sort out all the
components. Figure 6-3ashows a force of magnitude F
12.0 N applied to an 8.00 kg block at a downward angle of
u30.0. The coefficient of static friction between block
and floor is ms0.700; the coefficient of kinetic friction is
mk0.400. Does the block begin to slide or does it re-
main stationary? What is the magnitude of the frictional
force on the block?
KEY IDEAS
(1) When the object is stationary on a surface, the static fric-
tional force balances the force component that is attempting
to slide the object along the surface. (2) The maximum possi-
ble magnitude of that force is given by Eq. 6-1 ( fs,max msFN).
(3) If the component of the applied force along the surface
exceeds this limit on the static friction, the block begins to
slide. (4) If the object slides, the kinetic frictional force is
given by Eq. 6-2 ( fk mkFN).
Calculations: To see if the block slides (and thus to calcu-
late the magnitude of the frictional force), we must com-
pare the applied force component Fxwith the maximum
magnitude fs,max that the static friction can have. From the
triangle of components and full force shown in Fig. 6-3b,
we see that
FxFcos u
(12.0 N) cos 30 10.39 N. (6-3)
From Eq. 6-1, we know that fs,max msFN, but we need the
magnitude FNof the normal force to evaluate fs,max. Because
the normal force is vertical, we need to write Newton’s sec-
ond law (Fnet,ymay) for the vertical force components act-
ing on the block, as displayed in Fig. 6-3c. The gravitational
force with magnitude mg acts downward. The applied force
has a downward component FyFsin u. And the vertical
acceleration ayis just zero.Thus, we can write Newton’s sec-
F
y
x
u
(a)
(c)
Fg
Fy
FN
Block
Block
u
F
F
y
Fx
(b)
fsFx
(d)
Figure 6-3 (a) A force is applied to an initially stationary block. (b)
The components of the applied force. (c) The vertical force com-
ponents. (d) The horizontal force components.
Additional examples, video, and practice available at WileyPLUS
129
6-1 FRICTION
Inserting the initial speed v010.0 m/s, the final speed v0,
and the coefficient of kinetic friction mk0.60, we find that
the car’s stopping distance is
x x08.50 m 8.5 m. (Answer)
(b) What is the stopping distance if the road is covered with
ice with mk0.10?
Calculation: Our solution is perfectly fine through Eq. 6-12
but now we substitute this new mk, finding
xx051 m. (Answer)
Thus, a much longer clear path would be needed to avoid
the car hitting something along the way.
(c) Now let’s have the car sliding down an icy hill with an in-
clination of u5.00(a mild incline, nothing like the hills of
San Francisco). The free-body diagram shown in Fig. 6-4cis
like the ramp in Sample Problem 5.04 except, to be consis-
tent with Fig. 6-4b, the positive direction of the xaxis is
down the ramp.What now is the stopping distance?
Calculations: Switching from Fig.6-4bto cinvolves two ma-
jor changes. (1) Now a component of the gravitational force is
along the tilted xaxis, pulling the car down the hill. From
Sample Problem 5.04 and Fig. 5-15, that down-the-hill com-
ponent is mg sin u, which is in the positive direction of the x
axis in Fig. 6-4c. (2) The normal force (still perpendicular to
the road) now balances only a component of the gravitational
Sample Problem 6.02 Sliding to a stop on icy roads, horizontal and inclined
Some of the funniest videos on the web involve motorists
sliding uncontrollably on icy roads. Here let’s compare the
typical stopping distances for a car sliding to a stop from an
initial speed of 10.0 m/s on a dry horizontal road, an icy hori-
zontal road, and (everyone’s favorite) an icy hill.
(a) How far does the car take to slide to a stop on a hori-
zontal road (Fig. 6-4a) if the coefficient of kinetic friction is
mk0.60, which is typical of regular tires on dry pavement?
Let’s neglect any effect of the air on the car, assume that
the wheels lock up and the tires slide, and extend an xaxis
in the car’s direction of motion.
KEY IDEAS
(1) The car accelerates (its speed decreases) because a hori-
zontal frictional force acts against the motion, in the negative
direction of the xaxis. (2) The frictional force is a kinetic fric-
tional force with a magnitude given by Eq. 6-2 ( fkmkFN), in
which FNis the magnitude of the normal force on the car from
the road. (3) We can relate the frictional force to the resulting
acceleration by writing Newton’s second law (Fnet,xmax) for
motion along the road.
Calculations: Figure 6-4bshows the free-body diagram for the
car.The normal force is upward,the gravitational force is down-
ward, and the frictional force is horizontal. Because the fric-
tional force is the only force with an xcomponent, Newton’s
second law written for motion along the xaxis becomes
fkmax. (6-8)
Substituting fkmkFNgives us
mkFNmax. (6-9)
From Fig. 6-4bwe see that the upward normal force bal-
ances the downward gravitational force, so in Eq. 6-9 let’s
replace magnitude FNwith magnitude mg.Then we can can-
cel m(the stopping distance is thus independent of the car’s
mass—the car can be heavy or light, it does not matter).
Solving for axwe find
axmkg. (6-10)
Because this acceleration is constant, we can use the
constant-acceleration equations of Table 2-1. The easiest
choice for finding the sliding distance xx0is Eq. 2-16
which gives us
(6-11)
Substituting from Eq. 6-10, we then have
(6-12)xx0v2v2
0
2mkg.
xx0v2v2
0
2ax
.
(v2v2
02a(xx0)),
x – x0
v0
= 0.60
(a)
μ
v = 0
fk
Fg
FN
Car x
y
x
(b)
y
This is a free-body
diagram of the
forces on the car.
Frictional force
opposes the sliding.
Normal force
supports the car.
Gravitational force
pulls downward.
fk
FN
Fg
mg cos
(c)
u
u
u
mg sinu
Figure 6-4 (a) A car sliding to the right and finally stopping after
a displacement of 290 m. A free-body diagram for the car on
(b) a horizontal road and (c) a hill.
130 CHAPTER 6 FORCE AND MOTION—II
force, not the full force. From Sample Problem 5.04 (see Fig.
5-15i),we write that balance as
FNmg cos u.
In spite of these changes, we still want to write Newton’s
second law (Fnet,xmax) for the motion along the (now
tilted) xaxis.We have
fk mg sin umax,
mkFN mg sin umax,
and mkmg cos u mg sin umax.
Solving for the acceleration and substituting the given data
now give us
axmkgcos ugsin u
(0.10)(9.8 m/s2) cos 5.00(9.8 m/s2) sin 5.00
0.122 m/s2. (6-13)
Substituting this result into Eq. 6-11 gives us the stopping
distance hown the hill:
xx0409 m 400 m, (Answer)
which is about mi! Such icy hills separate people who can
do this calculation (and thus know to stay home) from peo-
ple who cannot (and thus end up in web videos).
1
4
Additional examples, video, and practice available at WileyPLUS
6-2 THE DRAG FORCE AND TERMINAL SPEED
After reading this module, you should be able to . . .
6.04 Apply the relationship between the drag force on an
object moving through air and the speed of the object.
6.05 Determine the terminal speed of an object falling
through air.
●When there is relative motion between air (or some other
fluid) and a body, the body experiences a drag force that
opposes the relative motion and points in the direction in
which the fluid flows relative to the body. The magnitude of
is related to the relative speed vby an experimentally deter-
mined drag coefficient Caccording to
,
where ris the fluid density (mass per unit volume) and A
is the effective cross-sectional area of the body (the area
D1
2C
Av2
D
:
D
:of a cross section taken perpendicular to the relative
velocity ).
●When a blunt object has fallen far enough through air, the
magnitudes of the drag force and the gravitational force
on the body become equal. The body then falls at a constant
terminal speed vtgiven by
.vtA2F
g
CrA
Fg
:
D
:
v
:
Learning Objectives
Key Ideas
The Drag Force and Terminal Speed
Afluid is anything that can flow—generally either a gas or a liquid.When there is
a relative velocity between a fluid and a body (either because the body moves
through the fluid or because the fluid moves past the body), the body experiences
adrag force that opposes the relative motion and points in the direction in
which the fluid flows relative to the body.
Here we examine only cases in which air is the fluid, the body is blunt (like
a baseball) rather than slender (like a javelin), and the relative motion is fast
enough so that the air becomes turbulent (breaks up into swirls) behind the
body. In such cases, the magnitude of the drag force is related to the relative
speed vby an experimentally determined drag coefficient Caccording to
(6-14)D1
2C
Av2,
D
:
D
:
where ris the air density (mass per volume) and Ais the effective cross-sectional
area of the body (the area of a cross section taken perpendicular to the
velocity ). The drag coefficient C(typical values range from 0.4 to 1.0) is not
truly a constant for a given body because if vvaries significantly, the value of C
can vary as well. Here,we ignore such complications.
Downhill speed skiers know well that drag depends on Aand v2. To reach
high speeds a skier must reduce Das much as possible by, for example, riding the
skis in the “egg position” (Fig.6-5) to minimize A.
Falling. When a blunt body falls from rest through air, the drag force is
directed upward; its magnitude gradually increases from zero as the speed of the
body increases.This upward force opposes the downward gravitational force
on the body. We can relate these forces to the body’s acceleration by writing
Newton’s second law for a vertical yaxis (Fnet,ymay) as
DFgma, (6-15)
where mis the mass of the body. As suggested in Fig. 6-6, if the body falls long
enough, Deventually equals Fg. From Eq. 6-15, this means that a0, and so the
body’s speed no longer increases. The body then falls at a constant speed, called
the terminal speed vt.
To find vt, we set a0 in Eq. 6-15 and substitute for Dfrom Eq. 6-14, obtaining
which gives (6-16)
Table 6-1 gives values of vtfor some common objects.
According to calculations* based on Eq. 6-14, a cat must fall about six
floors to reach terminal speed. Until it does so, FgDand the cat accelerates
downward because of the net downward force. Recall from Chapter 2
that your body is an accelerometer, not a speedometer. Because the cat also
senses the acceleration, it is frightened and keeps its feet underneath its body,
its head tucked in, and its spine bent upward, making Asmall, vtlarge, and in-
jury likely.
However, if the cat does reach vtduring a longer fall, the acceleration vanishes
and the cat relaxes somewhat, stretching its legs and neck horizontally outward and
vtA2F
g
C
A.
1
2C
Avt
2F
g0,
F
:
g
D
:
D
:
v
:
131
6-2 THE DRAG FORCE AND TERMINAL SPEED
Table 6-1 Some Terminal Speeds in Air
Object Terminal Speed (m/s) 95% Distancea(m)
Shot (from shot put) 145 2500
Sky diver (typical) 60 430
Baseball 42 210
Tennis ball 31 115
Basketball 20 47
Ping-Pong ball 9 10
Raindrop (radius 1.5 mm) 7 6
Parachutist (typical) 5 3
aThis is the distance through which the body must fall from rest to reach 95% of its terminal speed.
Based on Peter J. Brancazio,Sport Science, 1984, Simon & Schuster, New York.
Figure 6-5 This skier crouches in an “egg
position” so as to minimize her effective
cross-sectional area and thus minimize the
air drag acting on her.
Figure 6-6 The forces that act on a body
falling through air: (a) the body when it
has just begun to fall and (b) the free-
body diagram a little later, after a drag
force has developed. (c) The drag force
has increased until it balances the
gravitational force on the body. The body
now falls at its constant terminal speed.
Karl-Josef Hildenbrand/dpa/Landov LLC
Fg
(a)
Falling
body D
D
(b) (c)
Fg
Fg
As the cat's speed
increases, the upward
drag force increases
until it balances the
gravitational force.
*W. O. Whitney and C. J. Mehlhaff, “High-Rise Syndrome in Cats.” The Journal of the American
Veterinary Medical Association, 1987.
132 CHAPTER 6 FORCE AND MOTION—II
straightening its spine (it then resembles a flying squirrel). These actions increase
area Aand thus also, by Eq. 6-14, the drag D. The cat begins to slow because now
DFg(the net force is upward), until a new, smaller vtis reached. The decrease
in vtreduces the possibility of serious injury on landing. Just before the end of the
fall, when it sees it is nearing the ground, the cat pulls its legs back beneath its
body to prepare for the landing.
Humans often fall from great heights for the fun of skydiving. However, in
April 1987, during a jump, sky diver Gregory Robertson noticed that fellow
sky diver Debbie Williams had been knocked unconscious in a collision with
a third sky diver and was unable to open her parachute. Robertson, who
was well above Williams at the time and who had not yet opened his parachute
for the 4 km plunge, reoriented his body head-down so as to minimize Aand
maximize his downward speed. Reaching an estimated vtof 320 km/h, he
caught up with Williams and then went into a horizontal “spread eagle” (as in
Fig. 6-7) to increase Dso that he could grab her. He opened her parachute
and then, after releasing her, his own, a scant 10 s before impact. Williams
received extensive internal injuries due to her lack of control on landing but
survived.
Figure 6-7 Sky divers in a horizontal
“spread eagle” maximize air drag.
Steve Fitchett/Taxi/Getty Images
sity raand the water density rw,we obtain
(Answer)
Note that the height of the cloud does not enter into the
calculation.
(b) What would be the drop’s speed just before impact if
there were no drag force?
KEY IDEA
With no drag force to reduce the drop’s speed during the fall,
the drop would fall with the constant free-fall acceleration g,
so the constant-acceleration equations of Table 2-1 apply.
Calculation: Because we know the acceleration is g, the
initial velocity v0is 0, and the displacement xx0is h,we
use Eq. 2-16 to find v:
(Answer)
Had he known this, Shakespeare would scarcely have writ-
ten, “it droppeth as the gentle rain from heaven, upon the
place beneath.” In fact, the speed is close to that of a bullet
from a large-caliber handgun!
153 m/s 550 km/h.
v22gh 2(2)(9.8 m/s2)(1200 m)
7.4 m/s 27 km/h.
A(8)(1.5 103 m)(1000 kg/m3)(9.8 m/s2)
(3)(0.60)(1.2 kg/m3)
vtA2F
g
CraAA8pR3rwg
3Cra
R2A8Rrwg
3Cra
Sample Problem 6.03 Terminal speed of falling raindrop
A raindrop with radius R1.5 mm falls from a cloud that is
at height h1200 m above the ground.The drag coefficient
Cfor the drop is 0.60. Assume that the drop is spherical
throughout its fall. The density of water rwis 1000 kg/m3,
and the density of air rais 1.2 kg/m3.
(a) As Table 6-1 indicates, the raindrop reaches terminal
speed after falling just a few meters. What is the terminal
speed?
KEY IDEA
The drop reaches a terminal speed vtwhen the gravitational
force on it is balanced by the air drag force on it, so its accel-
eration is zero. We could then apply Newton’s second law
and the drag force equation to find vt, but Eq. 6-16 does all
that for us.
Calculations: To use Eq. 6-16, we need the drop’s effective
cross-sectional area Aand the magnitude Fgof the gravita-
tional force. Because the drop is spherical, Ais the area of a
circle (pR2) that has the same radius as the sphere. To find
Fg, we use three facts: (1) Fgmg, where mis the drop’s
mass; (2) the (spherical) drop’s volume is pR3; and
(3) the density of the water in the drop is the mass per vol-
ume, or rwm/V.Thus, we find
.
We next substitute this, the expression for A,and the given data
into Eq. 6-16. Being careful to distinguish between the air den-
F
gVrwg4
3pR3rwg
V4
3
Additional examples, video, and practice available at WileyPLUS
Uniform Circular Motion
From Module 4-5, recall that when a body moves in a circle (or a circular arc) at
constant speed v, it is said to be in uniform circular motion. Also recall that the
body has a centripetal acceleration (directed toward the center of the circle) of
constant magnitude given by
(centripetal acceleration), (6-17)
where Ris the radius of the circle. Here are two examples:
1. Rounding a curve in a car. You are sitting in the center of the rear seat of a car
moving at a constant high speed along a flat road. When the driver suddenly
turns left, rounding a corner in a circular arc, you slide across the seat toward the
right and then jam against the car wall for the rest of the turn.What is going on?
While the car moves in the circular arc, it is in uniform circular motion;
that is, it has an acceleration that is directed toward the center of the circle.
By Newton’s second law, a force must cause this acceleration. Moreover, the
force must also be directed toward the center of the circle. Thus, it is a cen-
tripetal force, where the adjective indicates the direction. In this example, the
centripetal force is a frictional force on the tires from the road; it makes the
turn possible.
If you are to move in uniform circular motion along with the car, there
must also be a centripetal force on you. However, apparently the frictional
force on you from the seat was not great enough to make you go in a circle
with the car. Thus, the seat slid beneath you, until the right wall of the car
jammed into you. Then its push on you provided the needed centripetal force
on you, and you joined the car’s uniform circular motion.
2. Orbiting Earth. This time you are a passenger in the space shuttle Atlantis. As
it and you orbit Earth, you float through your cabin.What is going on?
Both you and the shuttle are in uniform circular motion and have acceler-
ations directed toward the center of the circle.Again by Newton’s second law,
centripetal forces must cause these accelerations. This time the centripetal
forces are gravitational pulls (the pull on you and the pull on the shuttle) ex-
erted by Earth and directed radially inward, toward the center of Earth.
av2
R
133
6-3 UNIFORM CIRCULAR MOTION
6-3 UNIFORM CIRCULAR MOTION
After reading this module, you should be able to. . .
6.06 Sketch the path taken in uniform circular motion and
explain the velocity, acceleration, and force vectors
(magnitudes and directions) during the motion.
6.07 ldentify that unless there is a radially inward net force
(a centripetal force), an object cannot move in circular motion.
6.08 For a particle in uniform circular motion, apply the rela-
tionship between the radius of the path, the particle’s
speed and mass, and the net force acting on the particle.
●If a particle moves in a circle or a circular arc of radius Rat
constant speed v,the particle is said to be in uniform circular
motion. It then has a centripetal acceleration with magni-
tude given by
av2
R.
a
:
●This acceleration is due to a net centripetal force on the
particle, with magnitude given by
,
where mis the particle’s mass. The vector quantities and
are directed toward the center of curvature of the particle’s path.
F
:
a
:
Fmv2
R
Learning Objectives
Key Ideas
134 CHAPTER 6 FORCE AND MOTION—II
In both car and shuttle you are in uniform circular motion, acted on by a cen-
tripetal force—yet your sensations in the two situations are quite different. In
the car, jammed up against the wall, you are aware of being compressed by the
wall. In the orbiting shuttle, however, you are floating around with no sensation
of any force acting on you.Why this difference?
The difference is due to the nature of the two centripetal forces. In the
car, the centripetal force is the push on the part of your body touching the car
wall. You can sense the compression on that part of your body. In the shuttle,
the centripetal force is Earth’s gravitational pull on every atom of your body.
Thus, there is no compression (or pull) on any one part of your body and no
sensation of a force acting on you. (The sensation is said to be one of “weight-
lessness,” but that description is tricky. The pull on you by Earth has certainly
not disappeared and, in fact, is only a little less than it would be with you on
the ground.)
Another example of a centripetal force is shown in Fig. 6-8. There a hockey
puck moves around in a circle at constant speed vwhile tied to a string looped
around a central peg.This time the centripetal force is the radially inward pull on
the puck from the string.Without that force, the puck would slide off in a straight
line instead of moving in a circle.
Note again that a centripetal force is not a new kind of force.The name merely
indicates the direction of the force. It can, in fact, be a frictional force, a gravitational
force,the force from a car wall or a string, or any other force. For any situation:
Figure 6-8 An overhead view of a hockey puck moving with constant speed vin a
circular path of radius Ron a horizontal frictionless surface. The centripetal force on the
puck is , the pull from the string, directed inward along the radial axis rextending
through the puck.
T
:
String
Puck
R
vr
TThe puck moves
in uniform
circular motion
only because
of a toward-the-
center force.
A centripetal force accelerates a body by changing the direction of the body’s
velocity without changing the body’s speed.
From Newton’s second law and Eq. 6-17 (av2/R), we can write the magnitude
Fof a centripetal force (or a net centripetal force) as
(magnitude of centripetal force). (6-18)
Because the speed vhere is constant, the magnitudes of the acceleration and the
force are also constant.
However,the directions of the centripetal acceleration and force are not con-
stant; they vary continuously so as to always point toward the center of the circle.
For this reason, the force and acceleration vectors are sometimes drawn along a
radial axis rthat moves with the body and always extends from the center of the
circle to the body, as in Fig. 6-8. The positive direction of the axis is radially out-
ward, but the acceleration and force vectors point radially inward.
Fmv2
R
135
6-3 UNIFORM CIRCULAR MOTION
KEY IDEA
We can assume that Diavolo and his bicycle travel through
the top of the loop as a single particle in uniform circular
motion. Thus, at the top, the acceleration of this particlea
:
Sample Problem 6.04 Vertical circular loop, Diavolo
Largely because of riding in cars, you are used to horizon-
tal circular motion. Vertical circular motion would be a
novelty. In this sample problem, such motion seems to
defy the gravitational force.
In a 1901 circus performance, Allo “Dare Devil”
Diavolo introduced the stunt of riding a bicycle in a loop-
the-loop (Fig. 6-9a). Assuming that the loop is a circle with
radius R2.7 m, what is the least speed vthat Diavolo and
his bicycle could have at the top of the loop to remain in
contact with it there?
Figure 6-9 (a) Contemporary advertisement for Diavolo and
(b) free-body diagram for the performer at the top of the
loop.
y
Diavolo
and bicycle
a
Fg
FNThe net force
provides the
toward-the-center
acceleration.
The normal force
is from the
overhead loop.
(b)
(a)
Photograph reproduced with permission of
Circus World Museum
Additional examples, video, and practice available at WileyPLUS
Checkpoint 2
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats
mounted on a tall ring that rotates around a horizontal axis.When you ride in a
Ferris wheel at constant speed, what are the directions of your acceleration and the
normal force on you (from the always upright seat) as you pass through (a) the
highest point and (b) the lowest point of the ride? (c) How does the magnitude of
the acceleration at the highest point compare with that at the lowest point? (d) How
do the magnitudes of the normal force compare at those two points?
F
:
N
a
:
must have the magnitude av2/Rgiven by Eq. 6-17 and be
directed downward, toward the center of the circular loop.
Calculations: The forces on the particle when it is at the top
of the loop are shown in the free-body diagram of Fig 6-9b.
The gravitational force is downward along a yaxis; so is the
normal force on the particle from the loop (the loop can
push down, not pull up); so also is the centripetal acceleration
of the particle. Thus, Newton’s second law for ycomponents
(Fnet,ymay) gives us
FNFgm(a)
and (6-19)
If the particle has the least speed v needed to remain in
contact, then it is on the verge of losing contact with the loop
(falling away from the loop), which means that FN0 at the
top of the loop (the particle and loop touch but without any
normal force). Substituting 0 for FNin Eq. 6-19, solving for v,
and then substituting known values give us
(Answer)
Comments: Diavolo made certain that his speed at the
top of the loop was greater than 5.1 m/s so that he did not
lose contact with the loop and fall away from it. Note that
this speed requirement is independent of the mass of
Diavolo and his bicycle. Had he feasted on, say, pierogies
before his performance, he still would have had to exceed
only 5.1 m/s to maintain contact as he passed through the
top of the loop.
5.1 m/s.
v2gR 2(9.8 m/s2)(2.7 m)
F
Nmg m
v2
R
.
F
:
N
F
:
g
136 CHAPTER 6 FORCE AND MOTION—II
(b)
r
Car
Center fs
a
Fg
FN
y
R
(a)
r
FL
v
fs
The toward-the-
center force is
the frictional force.
Friction: toward the
center
Track-level view
of the forces
Normal force:
helps support car
Gravitational force:
pulls car downward
Negative lift: presses
car downward
Figure 6-10 (a) A race car moves around a flat curved track at constant speed v. The frictional
force provides the necessary centripetal force along a radial axis r. (b) A free-body diagram
(not to scale) for the car, in the vertical plane containing r.
f
:
s
Radial calculations: The frictional force is shown in the
free-body diagram of Fig. 6-10b. It is in the negative direc-
tion of a radial axis rthat always extends from the center of
curvature through the car as the car moves. The force pro-
duces a centripetal acceleration of magnitude v2/R. We can
relate the force and acceleration by writing Newton’s sec-
ond law for components along the raxis (Fnet,rmar) as
(6-20)
Substituting fs,max msFNfor fsleads us to
(6-21)
Vertical calculations: Next, let’s consider the vertical forces
on the car. The normal force is directed up, in the posi-
tive direction of the yaxis in Fig. 6-10b. The gravitational
force and the negative lift are directed down.
The acceleration of the car along the yaxis is zero. Thus we
can write Newton’s second law for components along the
yaxis (Fnet,ymay) as
FNmg FL0,
or FNmg FL. (6-22)
Combining results: Now we can combine our results along
the two axes by substituting Eq. 6-22 for FNin Eq. 6-21. Doing
so and then solving for FLlead to
(Answer)663.7 N 660 N.
(600 kg)
(28.6 m/s)2
(0.75)(100 m) 9.8 m/s2
FLm
v2
sRg
F
:
L
F
:
gmg
:
F
:
N
msFNm
v2
R
.
fsm
v2
R
.
f
:
s
Sample Problem 6.05 Car in flat circular turn
Upside-down racing: A modern race car is designed so that
the passing air pushes down on it, allowing the car to travel
much faster through a flat turn in a Grand Prix without fric-
tion failing. This downward push is called negative lift. Can a
race car have so much negative lift that it could be driven up-
side down on a long ceiling, as done fictionally by a sedan in
the first Men in Black movie?
Figure 6-10arepresents a Grand Prix race car of mass
m600 kg as it travels on a flat track in a circular arc of
radius R100 m. Because of the shape of the car and the
wings on it, the passing air exerts a negative lift down-
ward on the car. The coefficient of static friction between
the tires and the track is 0.75. (Assume that the forces on the
four tires are identical.)
(a) If the car is on the verge of sliding out of the turn when
its speed is 28.6 m/s, what is the magnitude of the negative
lift acting downward on the car?
KEY IDEAS
1. A centripetal force must act on the car because the car
is moving around a circular arc; that force must be
directed toward the center of curvature of the arc (here,
that is horizontally).
2. The only horizontal force acting on the car is a frictional
force on the tires from the road. So the required cen-
tripetal force is a frictional force.
3. Because the car is not sliding, the frictional force must
be a static frictional force (Fig.6-10a).
4. Because the car is on the verge of sliding, the magnitude
fsis equal to the maximum value fs,max msFN, where FN
is the magnitude of the normal force acting on the
car from the track.
F
:
N
f
:
s
F
:
L
F
:
L
137
6-3 UNIFORM CIRCULAR MOTION
Substituting our known negative lift of FL663.7 N and
solving for FL,90 give us
FL,90 6572 N 6600 N. (Answer)
Upside-down racing: The gravitational force is, of course,
the force to beat if there is a chance of racing upside down:
Fgmg (600 kg)(9.8 m/s2)
5880 N.
With the car upside down, the negative lift is an upward
force of 6600 N, which exceeds the downward 5880 N. Thus,
the car could run on a long ceiling provided that it moves at
about 90 m/s (324 km/h 201 mi/h). However, moving
that fast while right side up on a horizontal track is danger-
ous enough, so you are not likely to see upside-down racing
except in the movies.
(b) The magnitude FLof the negative lift on a car depends
on the square of the car’s speed v2, just as the drag force
does (Eq. 6-14). Thus, the negative lift on the car here is
greater when the car travels faster, as it does on a straight
section of track. What is the magnitude of the negative lift
for a speed of 90 m/s?
KEY IDEA
FLis proportional to v2.
Calculations: Thus we can write a ratio of the negative lift
FL,90 at v90 m/s to our result for the negative lift FLat v
28.6 m/s as
F
L,90
F
L
(90 m/s)2
(28.6 m/s)2.
of mass mas it moves at a constant speed vof 20 m/s around
a banked circular track of radius R190 m. (It is a normal
car, rather than a race car, which means that any vertical
force from the passing air is negligible.) If the frictional
force from the track is negligible, what bank angle upre-
vents sliding?
KEY IDEAS
Here the track is banked so as to tilt the normal force on
the car toward the center of the circle (Fig. 6-11b). Thus,
now has a centripetal component of magnitude FNr , directed
inward along a radial axis r. We want to find the value of
the bank angle usuch that this centripetal component
keeps the car on the circular track without need of friction.
F
:
N
F
:
N
Sample Problem 6.06 Car in banked circular turn
This problem is quite challenging in setting up but takes
only a few lines of algebra to solve. We deal with not only
uniformly circular motion but also a ramp. However, we will
not need a tilted coordinate system as with other ramps.
Instead we can take a freeze-frame of the motion and work
with simply horizontal and vertical axes. As always in this
chapter, the starting point will be to apply Newton’s second
law, but that will require us to identify the force component
that is responsible for the uniform circular motion.
Curved portions of highways are always banked (tilted)
to prevent cars from sliding off the highway. When a high-
way is dry, the frictional force between the tires and the road
surface may be enough to prevent sliding. When the high-
way is wet, however, the frictional force may be negligible,
and banking is then essential. Figure 6-11arepresents a car
(b)
y
r
FNr
R
(a)
θ
FNy
θ
v
rCar
Fg
FN
a
The toward-the-
center force is due
to the tilted track.
Track-level view
of the forces
The gravitational force
pulls car downward.
Tilted normal force
supports car and
provides the toward-
the-center force.
Figure 6-11 (a) A car moves around a curved banked road at constant speed v. The bank angle is exaggerated for clarity. (b)
A free-body diagram for the car, assuming that friction between tires and road is zero and that the car lacks negative lift.
The radially inward component FNr of the normal force (along radial axis r) provides the necessary centripetal force and
radial acceleration.
138 CHAPTER 6 FORCE AND MOTION—II
write Newton’s second law for components along the yaxis
(Fnet,ymay) as
FNcos umg m(0),
from which
FNcos umg. (6-24)
Combining results: Equation 6-24 also contains the
unknowns FNand m, but note that dividing Eq. 6-23 by
Eq. 6-24 neatly eliminates both those unknowns. Doing so,
replacing (sin u)/(cos u) with tan u, and solving for uthen
yield
. (Answer)tan1(20 m/s)2
(9.8 m/s2)(190 m) 12
tan1v2
gR
Radial calculation: As Fig. 6-11bshows (and as you
should verify), the angle that force makes with the ver-
tical is equal to the bank angle uof the track. Thus, the ra-
dial component FNr is equal to FNsin u. We can now write
Newton’s second law for components along the raxis
(Fnet,rmar) as
. (6-23)
We cannot solve this equation for the value of ubecause it
also contains the unknowns FNand m.
Vertical calculations: We next consider the forces and ac-
celeration along the yaxis in Fig. 6-11b. The vertical com-
ponent of the normal force is FNy FNcos u, the gravita-
tional force on the car has the magnitude mg, and the
acceleration of the car along the yaxis is zero. Thus we can
F
:
g
FN sin um
v2
R
F
:
N
Additional examples, video, and practice available at WileyPLUS
Friction When a force tends to slide a body along a surface, a
frictional force from the surface acts on the body.The frictional force
is parallel to the surface and directed so as to oppose the sliding. It is
due to bonding between the atoms on the body and the atoms on the
surface,an effect called cold-welding.
If the body does not slide, the frictional force is a static
frictional force . If there is sliding, the frictional force is a kinetic
frictional force .
1. If a body does not move, the static frictional force and the
component of parallel to the surface are equal in magnitude,
and is directed opposite that component. If the component
increases, fsalso increases.
2. The magnitude of has a maximum value fs,max given by
fs,max msFN, (6-1)
where msis the coefficient of static friction and FNis the magni-
tude of the normal force. If the component of parallel to the
surface exceeds fs,max, the static friction is overwhelmed and the
body slides on the surface.
3. If the body begins to slide on the surface, the magnitude of the
frictional force rapidly decreases to a constant value fkgiven
by
fkmkFN, (6-2)
where mkis the coefficient of kinetic friction.
Drag Force When there is relative motion between air (or
some other fluid) and a body, the body experiences a drag force
that opposes the relative motion and points in the direction in
which the fluid flows relative to the body. The magnitude of isD
:
D
:
F
:
f
:
s
f
:
s
F
:f
:
s
f
:
k
f
:
s
F
:
Review & Summary
related to the relative speed vby an experimentally determined
drag coefficient Caccording to
(6-14)
where ris the fluid density (mass per unit volume) and Ais the
effective cross-sectional area of the body (the area of a cross sec-
tion taken perpendicular to the relative velocity ).
Terminal Speed When a blunt object has fallen far enough
through air, the magnitudes of the drag force and the gravita-
tional force on the body become equal. The body then falls at a
constant terminal speed vtgiven by
(6-16)
Uniform Circular Motion If a particle moves in a circle or a
circular arc of radius Rat constant speed v, the particle is said to be
in uniform circular motion. It then has a centripetal acceleration
with magnitude given by
(6-17)
This acceleration is due to a net centripetal force on the particle,
with magnitude given by
(6-18)
where mis the particle’s mass. The vector quantities and are
directed toward the center of curvature of the particle’s path. A
particle can move in circular motion only if a net centripetal
force acts on it.
F
:
a
:
Fmv2
R,
av2
R.
a
:
vtA2F
g
CrA.
Fg
:D
:
v
:
D1
2C
Av2,
139
QUESTIONS
Questions
1In Fig. 6-12, if the box is station-
ary and the angle u between the hor-
izontal and force is increased
somewhat, do the following quanti-
ties increase, decrease, or remain the
same: (a) Fx; (b) fs; (c) FN; (d) fs,max? (e) If, instead, the box is sliding
and uis increased, does the magnitude of the frictional force on the
box increase, decrease, or remain the same?
2Repeat Question 1 for force angled upward instead of down-
ward as drawn.
3In Fig. 6-13, horizontal force
of magnitude 10 N is applied to a
box on a floor, but the box does not
slide. Then, as the magnitude of ver-
tical force is increased from zero,
do the following quantities increase,
decrease, or stay the same: (a) the magnitude of the frictional
force on the box; (b) the magnitude of the normal force on
the box from the floor; (c) the maximum value fs,max of the magni-
tude of the static frictional force on the box? (d) Does the box
eventually slide?
4In three experiments, three different horizontal forces are ap-
plied to the same block lying on the same countertop. The force
magnitudes are F112 N, F28 N, and F34 N. In each experi-
ment, the block remains stationary in spite of the applied force.
Rank the forces according to (a) the magnitude fsof the static fric-
tional force on the block from the countertop and (b) the maximum
value fs,max of that force,greatest first.
5If you press an apple crate against a wall so hard that the crate
cannot slide down the wall, what is the direction of (a) the static
frictional force on the crate from the wall and (b) the normal
force on the crate from the wall? If you increase your push,
what happens to (c) fs, (d) FN,and (e) fs,max?
6In Fig. 6-14, a block of mass mis held sta-
tionary on a ramp by the frictional force on
it from the ramp. A force , directed up the
ramp, is then applied to the block and grad-
ually increased in magnitude from zero.
During the increase, what happens to the di-
rection and magnitude of the frictional force
on the block?
7Reconsider Question 6 but with the force now directed
down the ramp. As the magnitude of is increased from zero,
what happens to the direction and magnitude of the frictional
force on the block?
8In Fig. 6-15, a horizontal force of 100 N is to be applied to a 10
kg slab that is initially stationary on a frictionless floor, to acceler-
ate the slab. A 10 kg block lies on top of the slab; the coefficient of
friction mbetween the block and the slab is not known, and the
F
:F
:
F
:
F
:
N
f
:
s
F
:
N
f
:
s
F
:
2
F
:
1
F
:
F
:
θ
x
F
Figure 6-12 Question 1.
block might slip. In fact, the contact between the block and the slab
might even be frictionless. (a) Considering that possibility, what is
the possible range of values for the magnitude of the slab’s acceler-
ation aslab? (Hint: You don’t need written calculations;just consider
extreme values for m.) (b) What is the possible range for the mag-
nitude ablock of the block’s acceleration?
9Figure 6-16 shows the overhead view of the path of an
amusement-park ride that travels at constant speed through five
circular arcs of radii R0,2R0, and 3R0. Rank the arcs according to
the magnitude of the centripetal force on a rider traveling in the
arcs, greatest first.
F2
F1
Figure 6-13 Question 3.
θ
F
Figure 6-14
Question 6.
100 N
Block
Slab
1
23
4
5
Figure 6-16 Question 9.
Figure 6-15 Question 8.
10 In 1987, as a Halloween stunt, two sky divers passed a
pumpkin back and forth between them while they were in free fall
just west of Chicago.The stunt was great fun until the last sky diver
with the pumpkin opened his parachute. The pumpkin broke free
from his grip, plummeted about 0.5 km, ripped through the roof of
a house, slammed into the kitchen floor, and splattered all over the
newly remodeled kitchen. From the sky diver’s viewpoint and from
the pumpkin’s viewpoint, why did the sky diver lose control of the
pumpkin?
11 A person riding a Ferris wheel moves through positions at
(1) the top, (2) the bottom, and (3) midheight. If the wheel rotates
at a constant rate, rank these three positions according to (a) the
magnitude of the person’s centripetal acceleration, (b) the magni-
tude of the net centripetal force on the person, and (c) the magni-
tude of the normal force on the person, greatest first.
12 During a routine flight in 1956, test pilot Tom Attridge put his
jet fighter into a 20dive for a test of the aircraft’s 20 mm machine
cannons. While traveling faster than sound at 4000 m altitude,
he shot a burst of rounds. Then, after allowing the cannons to cool,
he shot another burst at 2000 m; his speed was then 344 m/s, the
speed of the rounds relative to him was 730 m/s, and he was still in
a dive.
Almost immediately the canopy around him was shredded
and his right air intake was damaged. With little flying capability
left, the jet crashed into a wooded area, but Attridge managed to
escape the resulting explosion. Explain what apparently happened
just after the second burst of cannon rounds. (Attridge has been
the only pilot who has managed to shoot himself down.)
13 A box is on a ramp that is at angle uto the horizontal. As u
is increased from zero, and before the box slips, do the following
increase, decrease, or remain the same: (a) the component of the
gravitational force on the box, along the ramp, (b) the magnitude
of the static frictional force on the box from the ramp, (c) the
component of the gravitational force on the box, perpendicular
to the ramp, (d) the magnitude of the normal force on the
box from the ramp, and (e) the maximum value fs,max of the static
frictional force?
•9 A 3.5 kg block is pushed
along a horizontal floor by a force
of magnitude 15 N at an angle
40with the horizontal
(Fig. 6-19). The coefficient of ki-
netic friction between the block
and the floor is 0.25. Calculate the
magnitudes of (a) the frictional
force on the block from the floor
and (b) the block’s acceleration.
•10 Figure 6-20 shows an initially
stationary block of mass mon a
floor.A force of magnitude 0.500mg
is then applied at upward angle u
20.What is the magnitude of the ac-
celeration of the block across the
floor if the friction coefficients are (a) ms0.600 and mk0.500
and (b) ms0.400 and mk0.300?
•11 A 68 kg crate is dragged across a floor by pulling on
a rope attached to the crate and inclined 158above the horizontal.
(a) If the coefficient of static friction is 0.50, what minimum force
magnitude is required from the rope to start the crate moving?
(b) If mk0.35, what is the magnitude of the initial acceleration of
the crate?
•12 In about 1915, Henry Sincosky of Philadelphia suspended
himself from a rafter by gripping the rafter with the thumb of each
SSM
u
F
:
140 CHAPTER 6 FORCE AND MOTION—II
Module 6-1 Friction
•1 The floor of a railroad flatcar is loaded with loose crates hav-
ing a coefficient of static friction of 0.25 with the floor. If the train
is initially moving at a speed of 48 km/h, in how short a distance
can the train be stopped at constant acceleration without causing
the crates to slide over the floor?
•2 In a pickup game of dorm shuffleboard, students crazed by fi-
nal exams use a broom to propel a calculus book along the dorm
hallway. If the 3.5 kg book is pushed from rest through a distance
of 0.90 m by the horizontal 25 N force from the broom and then
has a speed of 1.60 m/s, what is the coefficient of kinetic friction be-
tween the book and floor?
•3 A bedroom bureau with a mass of 45 kg, includ-
ing drawers and clothing, rests on the floor. (a) If the coefficient of
static friction between the bureau and the floor is 0.45, what is the
magnitude of the minimum horizontal force that a person must ap-
ply to start the bureau moving? (b) If the drawers and clothing,
with 17 kg mass, are removed before the bureau is pushed, what is
the new minimum magnitude?
•4 A slide-loving pig slides down a certain 35slide in twice the
time it would take to slide down a frictionless 35slide.What is the
coefficient of kinetic friction between the pig and the slide?
•5 A 2.5 kg block is initially at rest on a horizontal surface. A
WWWSSM
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
P
F
Figure 6-17 Problem 5.
F
x
y
θ
Figure 6-20 Problem 10.
θ
F
Figure 6-19
Problems 9 and 32.
Figure 6-18 Problem 8. What moved the stone?
Jerry Schad/Photo Researchers, Inc.
•6 A baseball player with mass m79 kg, sliding into second
base, is retarded by a frictional force of magnitude 470 N. What is
the coefficient of kinetic friction mkbetween the player and the
ground?
•7 A person pushes horizontally with a force of 220 N
on a 55 kg crate to move it across a level floor. The coefficient
of kinetic friction between the crate and the floor is 0.35. What is
the magnitude of (a) the frictional force and (b) the acceleration of
the crate?
•8 The mysterious sliding stones. Along the remote
Racetrack Playa in Death Valley, California, stones sometimes
gouge out prominent trails in the desert floor, as if the stones
had been migrating (Fig. 6-18). For years curiosity mounted
about why the stones moved. One explanation was that strong
winds during occasional rainstorms would drag the rough stones
ILWSSM
horizontal force of magnitude 6.0 N and a vertical force are
then applied to the block (Fig. 6-17). The coefficients of friction for
the block and surface are ms0.40 and mk0.25. Determine the
magnitude of the frictional force acting on the block if the magni-
tude of is (a) 8.0 N, (b) 10 N,and (c) 12 N.P
:
P
:
F
:
over ground softened by rain. When the desert dried out, the
trails behind the stones were hard-baked in place. According to
measurements, the coefficient of kinetic friction between the
stones and the wet playa ground is about 0.80. What horizontal
force must act on a 20 kg stone (a typical mass) to maintain the
stone’s motion once a gust has started it moving? (Story contin-
ues with Problem 37.)
Cheerios (mass mC1.0 kg) and a
box of Wheaties (mass mW3.0
kg) are accelerated across a hori-
zontal surface by a horizontal force
applied to the Cheerios box. The
magnitude of the frictional force on the Cheerios box is 2.0 N,
and the magnitude of the frictional force on the Wheaties box is
4.0 N. If the magnitude of is 12 N, what is the magnitude of the
force on the Wheaties box from the Cheerios box?
••21 An initially stationary box of sand is to be pulled across a
floor by means of a cable in which the tension should not exceed
1100 N. The coefficient of static friction between the box and the
floor is 0.35. (a) What should be the angle between the cable and
the horizontal in order to pull the greatest possible amount of sand,
and (b) what is the weight of the sand and box in that situation?
••22 In Fig. 6-23, a sled is held on an inclined plane by a cord
pulling directly up the plane. The sled is to be on the verge of
moving up the plane. In Fig. 6-
28, the magnitude Frequired of
the cord’s force on the sled is
plotted versus a range of values
for the coefficient of static fric-
tion msbetween sled and plane:
F12.0 N, F25.0 N, and m2
0.50.At what angle uis the plane
inclined?
F
:
F
:
141
PROBLEMS
initially at rest on a plane inclined at angle u15to the horizon-
tal. The positive direction of the xaxis is up the plane. Between
block and plane, the coefficient of static friction is ms0.50 and
the coefficient of kinetic friction is mk0.34. In unit-vector nota-
tion, what is the frictional force on the block from the plane when
is (a) (5.0 N) , (b) (8.0 N) , and (c) (15 N) ?
••18 You testify as an expert witness in a case involving an acci-
dent in which car Aslid into the rear of car B, which was stopped at
a red light along a road headed down a hill (Fig. 6-25). You find
that the slope of the hill is u12.0, that the cars were separated
by distance d24.0 m when the driver of car Aput the car into a
slide (it lacked any automatic anti-brake-lock system), and that the
speed of car Aat the onset of braking was v018.0 m/s.With what
speed did car Ahit car Bif the coefficient of kinetic friction was
(a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with
wet leaves)?
i
ˆ
i
ˆ
i
ˆ
P
:
hand on one side and the fingers on the opposite
side (Fig. 6-21). Sincosky’s mass was 79 kg. If the
coefficient of static friction between hand and
rafter was 0.70, what was the least magnitude of
the normal force on the rafter from each thumb
or opposite fingers? (After suspending himself,
Sincosky chinned himself on the rafter and then
moved hand-over-hand along the rafter. If you do
not think Sincosky’s grip was remarkable, try to
repeat his stunt.)
•13 A worker pushes horizontally on a 35 kg
crate with a force of magnitude 110 N. The coeffi-
cient of static friction between the crate and the
floor is 0.37. (a) What is the value of fs,max under
the circumstances? (b) Does the crate move?
(c) What is the frictional force on the crate from
the floor? (d) Suppose, next, that a second worker
pulls directly upward on the crate to help out.
What is the least vertical pull that will allow the
first worker’s 110 N push to move the crate? (e)
If, instead, the second worker pulls horizontally to help out, what is
the least pull that will get the crate moving?
•14 Figure 6-22 shows the cross
section of a road cut into the side of
a mountain. The solid line AArep-
resents a weak bedding plane along
which sliding is possible. Block B
directly above the highway is sepa-
rated from uphill rock by a large
crack (called a joint), so that only
friction between the block and the
bedding plane prevents sliding. The
mass of the block is 1.8 107kg, the dip angle uof the bedding
plane is 24, and the coefficient of static friction between block
and plane is 0.63. (a) Show that the block will not slide under
these circumstances. (b) Next, water seeps into the joint and ex-
pands upon freezing, exerting on the block a force parallel to
AA. What minimum value of force magnitude Fwill trigger a
slide down the plane?
•15 The coefficient of static friction between Teflon and scram-
bled eggs is about 0.04. What is the smallest angle from the hori-
zontal that will cause the eggs to slide across the bottom of a
Teflon-coated skillet?
••16 A loaded penguin sled weigh-
ing 80 N rests on a plane inclined at
angle u20to the horizontal (Fig.
6-23). Between the sled and the
plane, the coefficient of static
friction is 0.25, and the coefficient of
kinetic friction is 0.15. (a) What is
the least magnitude of the force
parallel to the plane, that will pre-
vent the sled from slipping down the plane? (b) What is the mini-
mum magnitude Fthat will start the sled moving up the plane? (c)
What value of Fis required to
move the sled up the plane at con-
stant velocity?
••17 In Fig. 6-24, a force acts on
a block weighing 45 N. The block is
P
:
F
:
,
F
:
Joint with ice
θ
A
A'
B
F
Figure 6-22 Problem 14.
Figure 6-21
Problem 12.
θ
B
A
d
v0
Figure 6-25 Problem 18.
θ
P
x
Figure 6-24 Problem 17.
F
θ
Figure 6-23
Problems 16 and 22.
••19 A 12 N horizontal force
pushes a block weighing 5.0 N
against a vertical wall (Fig. 6-26).
The coefficient of static friction be-
tween the wall and the block is 0.60,
and the coefficient of kinetic friction
is 0.40. Assume that the block is not
moving initially. (a) Will the block move? (b) In unit-vector nota-
tion, what is the force on the block from the wall?
••20 In Fig. 6-27, a box of
F
:
mC
FmW
Figure 6-27 Problem 20.
x
y
F
Figure 6-26 Problem 19.
F
F
2
F
1
0
μ
2
μ
s
Figure 6-28 Problem 22.
142 CHAPTER 6 FORCE AND MOTION—II
a
a
1
–a
1
μ
0
k3
μ
k2
μ
k
Figure 6-36 Problem 32.
••30 A toy chest and its contents have a combined weight of
180 N. The coefficient of static friction between toy chest and floor
is 0.42. The child in Fig. 6-35 attempts to move the chest across the
floor by pulling on an attached rope. (a) If uis 42, what is the mag-
nitude of the force that the child must exert on the rope to put
the chest on the verge of moving? (b) Write an expression for the
magnitude Frequired to put the chest on the verge of moving as a
function of the angle u. Determine (c) the value of ufor which Fis
a minimum and (d) that minimum magnitude.
F
:
••23 When the three blocks in
Fig. 6-29 are released from rest, they
accelerate with a magnitude of
0.500 m/s2. Block 1 has mass M,
block 2 has 2M, and block 3 has 2M.
What is the coefficient of kinetic
friction between block 2 and the
table?
••24 A 4.10 kg block is pushed
along a floor by a constant applied
force that is horizontal and has a
magnitude of 40.0 N. Figure 6-30
gives the block’s speed vversus
time tas the block moves along an x
axis on the floor. The scale of the
figure’s vertical axis is set by vs
5.0 m/s. What is the coefficient of
kinetic friction between the block
and the floor?
••25 Block Bin Fig.
6-31 weighs 711 N. The coefficient of
static friction between block and
table is 0.25; angle uis 30; assume
that the cord between Band the
knot is horizontal. Find the maxi-
mum weight of block Afor which
the system will be stationary.
••26 Figure 6-32 shows three
crates being pushed over a concrete
floor by a horizontal force of
magnitude 440 N. The masses of the
crates are m130.0 kg, m210.0
kg, and m320.0 kg.The coefficient
of kinetic friction between the floor
and each of the crates is 0.700. (a)
What is the magnitude F32 of the
force on crate 3 from crate 2? (b) If
the crates then slide onto a polished
floor, where the coefficient of kinetic friction is less than 0.700, is
magnitude F32 more than, less than, or the same as it was when the
coefficient was 0.700?
••27 Body Ain Fig. 6-33 weighs
102 N, and body Bweighs 32 N. The
coefficients of friction between A
and the incline are ms0.56 and
mk0.25. Angle uis 40. Let the
positive direction of an xaxis be up
the incline. In unit-vector notation,
what is the acceleration of Aif Ais
initially (a) at rest, (b) moving up
the incline, and (c) moving down
the incline?
••28 In Fig. 6-33, two blocks are connected over a pulley. The
mass of block Ais 10 kg, and the coefficient of kinetic friction be-
tween Aand the incline is 0.20. Angle uof the incline is 30. Block
Aslides down the incline at constant speed. What is the mass of
block B? Assume the connecting rope has negligible mass. (The
pulley’s function is only to redirect the rope.)
F
:
WWWSSM
••29 In Fig. 6-34, blocks Aand Bhave weights of 44 N and 22
N, respectively. (a) Determine the minimum weight of block Cto
keep Afrom sliding if msbetween Aand the table is 0.20. (b) Block
Csuddenly is lifted off A. What is the acceleration of block Aif mk
between Aand the table is 0.15?
3
2
1
Figure 6-29 Problem 23.
v (m/s)
0 0.5
t
(
s
)
1.0
vs
Figure 6-30 Problem 24.
B
A
Knot
θ
Figure 6-31 Problem 25.
m1
m2m3
F
Figure 6-32 Problem 26.
Frictionless,
massless pulley
A
B
θ
Figure 6-33
Problems 27 and 28.
Figure 6-34 Problem 29.
Frictionless,
massless pulley
B
C
A
Figure 6-35 Problem 30.
θ
by a massless string and slide down a 30inclined plane.The coeffi-
cient of kinetic friction between the lighter block and the plane is
0.10, and the coefficient between the heavier block and the plane is
0.20. Assuming that the lighter block leads, find (a) the magnitude
of the acceleration of the blocks and (b) the tension in the taut
string.
••32 A block is pushed across a floor by a constant force that is
applied at downward angle u(Fig. 6-19). Figure 6-36 gives the accel-
eration magnitude aversus a range of values for the coefficient of
kinetic friction mkbetween block and floor: a13.0 m/s2,mk2
0.20, and mk30.40.What is the value of u?
••31 Two blocks, of weights 3.6 N and 7.2 N, are connected
SSM
kg and M88 kg) in Fig. 6-38 are
not attached to each other.The coef-
ficient of static friction between the
blocks is ms0.38, but the surface
beneath the larger block is friction-
less. What is the minimum magnitude
of the horizontal force required to
keep the smaller block from slipping down the larger block?
Module 6-2 The Drag Force and Terminal Speed
•36 The terminal speed of a sky diver is 160 km/h in the spread-
eagle position and 310 km/h in the nosedive position. Assuming
that the diver’s drag coefficient Cdoes not change from one posi-
tion to the other, find the ratio of the effective cross-sectional area
Ain the slower position to that in the faster position.
••37 Continuation of Problem 8. Now assume that
Eq. 6-14 gives the magnitude of the air drag force on the typical
20 kg stone, which presents to the wind a vertical cross-sectional
area of 0.040 m2and has a drag coefficient Cof 0.80. Take the air
density to be 1.21 kg/m3, and the coefficient of kinetic friction to
be 0.80. (a) In kilometers per hour, what wind speed Valong the
ground is needed to maintain the stone’s motion once it has
started moving? Because winds along the ground are retarded by
the ground, the wind speeds reported for storms are often meas-
ured at a height of 10 m. Assume wind speeds are 2.00 times
those along the ground. (b) For your answer to (a), what wind
speed would be reported for the storm? (c) Is that value reason-
able for a high-speed wind in a storm? (Story continues with
Problem 65.)
••38 Assume Eq. 6-14 gives the drag force on a pilot plus ejection
seat just after they are ejected from a plane traveling horizontally
at 1300 km/h.Assume also that the mass of the seat is equal to the
mass of the pilot and that the drag coefficient is that of a sky diver.
Making a reasonable guess of the pilot’s mass and using the
appropriate vtvalue from Table 6-1, estimate the magnitudes of
(a) the drag force on the pilot seat and (b) their horizontal de-
celeration (in terms of g), both just after ejection. (The result of
(a) should indicate an engineering requirement: The seat must in-
clude a protective barrier to deflect the initial wind blast away
from the pilot’s head.)
••39 Calculate the ratio of the drag force on a jet flying at
1000 km/h at an altitude of 10 km to the drag force on a prop-
driven transport flying at half that speed and altitude. The density
F
:
•••34 In Fig. 6-37, a slab of mass
m140 kg rests on a frictionless
floor, and a block of mass m210
kg rests on top of the slab. Between
block and slab, the coefficient of
static friction is 0.60, and the coefficient of kinetic friction is 0.40.A
horizontal force of magnitude 100 N begins to pull directly on
the block, as shown. In unit-vector notation, what are the resulting
accelerations of (a) the block and (b) the slab?
•••35 The two blocks (m16
ILW
F
:
•••33 A 1000 kg boat is traveling at 90 km/h when its engine
is shut off. The magnitude of the frictional force between boat
and water is proportional to the speed vof the boat: fk70v, where
vis in meters per second and fkis in newtons. Find the time required
for the boat to slow to 45 km/h.
f
:
k
SSM
143
PROBLEMS
of air is 0.38 kg/m3at 10 km and 0.67 kg/m3at 5.0 km. Assume that
the airplanes have the same effective cross-sectional area and drag
coefficient C.
••40 In downhill speed skiing a skier is retarded by both
the air drag force on the body and the kinetic frictional force on the
skis. (a) Suppose the slope angle is u40.0, the snow is dry snow
with a coefficient of kinetic friction mk0.0400, the mass of the
skier and equipment is m85.0 kg, the cross-sectional area of the
(tucked) skier is A1.30 m2, the drag coefficient is C0.150, and
the air density is 1.20 kg/m3. (a) What is the terminal speed? (b) If a
skier can vary Cby a slight amount dC by adjusting, say, the hand
positions, what is the corresponding variation in the terminal
speed?
Module 6-3 Uniform Circular Motion
•41 A cat dozes on a stationary merry-go-round in an amuse-
ment park, at a radius of 5.4 m from the center of the ride.Then the
operator turns on the ride and brings it up to its proper turning
rate of one complete rotation every 6.0 s. What is the least coeffi-
cient of static friction between the cat and the merry-go-round that
will allow the cat to stay in place, without sliding (or the cat cling-
ing with its claws)?
•42 Suppose the coefficient of static friction between the road
and the tires on a car is 0.60 and the car has no negative lift. What
speed will put the car on the verge of sliding as it rounds a level
curve of 30.5 m radius?
•43 What is the smallest radius of an unbanked (flat) track
around which a bicyclist can travel if her speed is 29 km/h and the
msbetween tires and track is 0.32?
•44 During an Olympic bobsled run, the Jamaican team makes a
turn of radius 7.6 m at a speed of 96.6 km/h. What is their accelera-
tion in terms of g?
••45 A student of weight 667 N rides a
steadily rotating Ferris wheel (the student sits upright). At the
highest point, the magnitude of the normal force on the student
from the seat is 556 N. (a) Does the student feel “light” or “heavy”
there? (b) What is the magnitude of at the lowest point? If the
wheel’s speed is doubled, what is the magnitude FNat the (c) high-
est and (d) lowest point?
••46 A police officer in hot pursuit drives her car through a circular
turn of radius 300 m with a constant speed of 80.0 km/h. Her mass is
55.0 kg.What are (a) the magnitude and (b) the angle (relative to ver-
tical) of the net force of the officer on the car seat? (Hint: Consider
both horizontal and vertical forces.)
••47 A circular-motion addict of mass 80 kg rides a Ferris
wheel around in a vertical circle of radius 10 m at a constant speed
of 6.1 m/s. (a) What is the period of the motion? What is the mag-
nitude of the normal force on the addict from the seat when both
go through (b) the highest point of the circular path and (c) the
lowest point?
••48 A roller-coaster car at an amusement park has a mass
of 1200 kg when fully loaded with passengers. As the car passes
over the top of a circular hill of radius 18 m, assume that its speed
is not changing. At the top of the hill, what are the (a) magnitude
FNand (b) direction (up or down) of the normal force on the car
from the track if the car’s speed is v11 m/s? What are (c) FNand
(d) the direction if v14 m/s?
F
:
N
F
:
N
ILWSSM
ILW
Figure 6-37 Problem 34.
m2
m1
x
= 0
F
μ
Frictionless
m
M
F
Figure 6-38 Problem 35.
••58 Brake or turn? Figure 6-
44 depicts an overhead view of a car’s
path as the car travels toward a wall.
Assume that the driver begins to
brake the car when the distance to
the wall is d107 m, and take the
car’s mass as m1400 kg, its initial
speed as v035 m/s, and the coeffi-
cient of static friction as ms0.50.
Assume that the car’s weight is dis-
tributed evenly on the four wheels,
even during braking. (a) What magni-
tude of static friction is needed (between tires and road) to stop
the car just as it reaches the wall? (b) What is the maximum pos-
sible static friction fs,max? (c) If the coefficient of kinetic friction
between the (sliding) tires and the road is mk0.40, at what
speed will the car hit the wall? To avoid the crash, a driver could
elect to turn the car so that it just barely misses the wall, as
shown in the figure. (d) What magnitude of frictional force would
be required to keep the car in a circular path of radius dand at
the given speed v0, so that the car moves in a quarter circle and
then parallel to the wall? (e) Is the required force less than fs,max
so that a circular path is possible?
144 CHAPTER 6 FORCE AND MOTION—II
F
v
(a)
F
T
(b)
Figure 6-40 Problem 50.
dv in the speed with rheld constant, and (c) a variation dT in the
period with rheld constant?
••55 A bolt is threaded onto one
end of a thin horizontal rod, and
the rod is then rotated horizontally
about its other end. An engineer
monitors the motion by flashing a
strobe lamp onto the rod and bolt,
adjusting the strobe rate until the
bolt appears to be in the same
eight places during each full rota-
tion of the rod (Fig. 6-42). The strobe rate is 2000 flashes per sec-
ond; the bolt has mass 30 g and is at radius 3.5 cm. What is the
magnitude of the force on the bolt from the rod?
••56 A banked circular highway curve is designed for traffic
moving at 60 km/h. The radius of the curve is 200 m. Traffic is
moving along the highway at 40 km/h on a rainy day. What is the
minimum coefficient of friction between tires and road that will
allow cars to take the turn without sliding off the road? (Assume
the cars do not have negative lift.)
••57 A puck of mass m1.50 kg slides in a circle of radius
r20.0 cm on a frictionless table while attached to a hanging
cylinder of mass M2.50 kg by means of a cord that extends
through a hole in the table (Fig. 6-43).What speed keeps the cylin-
der at rest?
••49 In Fig. 6-39, a car is driven at constant speed over a circu-
lar hill and then into a circular valley with the same radius. At the
top of the hill, the normal force on the driver from the car seat is 0.
The driver’s mass is 70.0 kg. What is the magnitude of the normal
force on the driver from the seat when the car passes through the
bottom of the valley?
Radius
Radius
Figure 6-39 Problem 49.
••50 An 85.0 kg passenger is made to move along a circular path
of radius r3.50 m in uniform circular motion. (a) Figure 6-40ais
a plot of the required magnitude Fof the net centripetal force for a
range of possible values of the passenger’s speed v. What is the
plot’s slope at v8.30 m/s? (b) Figure 6-40bis a plot of Ffor a
range of possible values of T, the period of the motion.What is the
plot’s slope at T2.50 s?
••51 An airplane is fly-
ing in a horizontal circle at a speed of
480 km/h (Fig. 6-41). If its wings are
tilted at angle u 40to the horizon-
tal, what is the radius of the circle in
which the plane is flying? Assume
that the required force is provided
entirely by an “aerodynamic lift” that
is perpendicular to the wing surface.
••52 An amusement park
ride consists of a car moving in a ver-
tical circle on the end of a rigid boom
of negligible mass. The combined weight of the car and riders is 5.0
kN, and the circle’s radius is 10 m. At the top of the circle, what
are the (a) magnitude FBand (b) direction (up or down) of
the force on the car from the boom if the car’s speed is v5.0 m/s?
What are (c) FBand (d) the direction if v12 m/s?
••53 An old streetcar rounds a flat corner of radius 9.1 m, at
16 km/h. What angle with the vertical will be made by the loosely
hanging hand straps?
••54 In designing circular rides for amusement parks,
mechanical engineers must consider how small variations in cer-
tain parameters can alter the net force on a passenger. Consider a
passenger of mass mriding around a horizontal circle of radius rat
speed v. What is the variation dF in the net force magnitude for
(a) a variation dr in the radius with vheld constant, (b) a variation
WWWSSM
θ
Figure 6-41 Problem 51.
Strobed
positions
Bolt
Rod
Figure 6-42 Problem 55.
m
r
M
m
r
M
Figure 6-43 Problem 57.
Car path
Wall
d
Figure 6-44
Problem 58.
145
PROBLEMS
•••59 In Fig. 6-45, a 1.34 kg
ball is connected by means of two mass-
less strings, each of length L1.70 m, to
a vertical, rotating rod.The strings are tied
to the rod with separation d1.70 m and
are taut. The tension in the upper string is
35 N.What are the (a) tension in the lower
string, (b) magnitude of the net force
on the ball, and (c) speed of the ball? (d)
What is the direction of ?F
:
net
F
:
net
ILWSSM 63 In Fig. 6-49, a 49 kg rock climber is climbing a “chim-
ney.” The coefficient of static friction between her shoes and the
rock is 1.2; between her back and the rock is 0.80. She has reduced
her push against the rock until her back and her shoes are on the
verge of slipping. (a) Draw a free-body diagram of her. (b) What is
the magnitude of her push against the rock? (c) What fraction of
her weight is supported by the frictional force on her shoes?
m1
m2
θ
Figure 6-46 Problem 60.
61 A block of mass mt4.0 kg is put on top of a block of
mass mb5.0 kg.To cause the top block to slip on the bottom one
while the bottom one is held fixed, a horizontal force of at least 12
N must be applied to the top block. The assembly of blocks is now
placed on a horizontal, frictionless table (Fig. 6-47). Find the mag-
nitudes of (a) the maximum horizontal force that can be applied
to the lower block so that the blocks will move together and (b) the
resulting acceleration of the blocks.
F
:
SSM
mt
F
mb
Figure 6-47 Problem 61.
62 A 5.00 kg stone is rubbed across the horizontal ceiling of a
cave passageway (Fig. 6-48). If the coefficient of kinetic friction is
0.65 and the force applied to the stone is angled at u70.0°, what
must the magnitude of the force be for the stone to move at constant
velocity?
Stone
θ
F
Figure 6-48 Problem 62.
Figure 6-49 Problem 63.
64 A high-speed railway car goes around a flat, horizontal circle
of radius 470 m at a constant speed. The magnitudes of the hori-
zontal and vertical components of the force of the car on a 51.0 kg
passenger are 210 N and 500 N, respectively. (a) What is the magni-
tude of the net force (of all the forces) on the passenger? (b) What
is the speed of the car?
65 Continuation of Problems 8 and 37. Another explana-
tion is that the stones move only when the water dumped on the
playa during a storm freezes into a large, thin sheet of ice. The
stones are trapped in place in the ice. Then, as air flows across
the ice during a wind, the air-drag forces on the ice and stones
move them both, with the stones gouging out the trails.The magni-
tude of the air-drag force on this horizontal “ice sail” is given by
Dice 4CicerAicev2, where Cice is the drag coefficient (2.0 103), r
is the air density (1.21 kg/m3), Aice is the horizontal area of the ice,
and vis the wind speed along the ice.
Assume the following:The ice sheet measures 400 m by 500 m
by 4.0 mm and has a coefficient of kinetic friction of 0.10 with the
ground and a density of 917 kg/m3. Also assume that 100 stones
identical to the one in Problem 8 are trapped in the ice. To main-
tain the motion of the sheet, what are the required wind speeds (a)
near the sheet and (b) at a height of 10 m? (c) Are these reason-
able values for high-speed winds in a storm?
66 In Fig. 6-50, block 1 of mass m12.0 kg and block 2 of
mass m23.0 kg are connected by a string of negligible mass and
are initially held in place. Block 2 is on a frictionless surface tilted
at u30. The coefficient of kinetic friction between block 1 and
the horizontal surface is 0.25. The pulley has negligible mass and
friction. Once they are released, the blocks move. What then is the
tension in the string?
θ
m1
m2
Figure 6-50 Problem 66.
Additional Problems
60 In Fig. 6-46, a box of ant aunts (total
mass m11.65 kg) and a box of ant un-
cles (total mass m23.30 kg) slide down an inclined plane while
attached by a massless rod parallel to the plane. The angle of in-
cline is u30.0°. The coefficient of kinetic friction between the
aunt box and the incline is m10.226; that between the uncle box
and the incline is m20.113. Compute (a) the tension in the rod
and (b) the magnitude of the common acceleration of the two
boxes. (c) How would the answers to (a) and (b) change if the un-
cles trailed the aunts?
d
Rotating rod
L
L
Figure 6-45
Problem 59.
146 CHAPTER 6 FORCE AND MOTION—II
θ
P
Figure 6-52 Problem 69.
φ
New slope
Original slope
θ
Figure 6-55 Problem 76.
67 In Fig. 6-51, a crate slides down an inclined right-angled
trough.The coefficient of kinetic friction between the crate and the
trough is mk. What is the acceleration of the crate in terms of mk,u,
and g?
To three significant figures, what is the magnitude of that applied
force if it puts the block on the verge of sliding when the force is
directed (a) horizontally, (b) upward at 60.0from the horizontal,
and (c) downward at 60.0° from the horizontal?
72 A box of canned goods slides down a ramp from street level
into the basement of a grocery store with acceleration 0.75 m/s2di-
rected down the ramp. The ramp makes an angle of 40with the
horizontal. What is the coefficient of kinetic friction between the
box and the ramp?
73 In Fig. 6-54, the coefficient of kinetic friction
between the block and inclined plane is 0.20, and
angle uis 60. What are the (a) magnitude aand
(b) direction (up or down the plane) of the block’s
acceleration if the block is sliding down the plane?
What are (c) aand (d) the direction if the block is
sent sliding up the plane?
74 A 110 g hockey puck sent sliding over ice is
stopped in 15 m by the frictional force on it from the ice. (a) If its ini-
tial speed is 6.0 m/s, what is the magnitude of the frictional force? (b)
What is the coefficient of friction between the puck and the ice?
75 A locomotive accelerates a 25-car train along a level track.
Every car has a mass of 5.0 104kg and is subject to a frictional
force f250v, where the speed vis in meters per second and the
force fis in newtons. At the instant when the speed of the train is
30 km/h, the magnitude of its acceleration is 0.20 m/s2. (a) What
is the tension in the coupling between the first car and the
locomotive? (b) If this tension is equal to the maximum force the
locomotive can exert on the train, what is the steepest grade up
which the locomotive can pull the train at 30 km/h?
76 A house is built on the top of a hill with a nearby slope at angle
u45(Fig. 6-55). An engineering study indicates that the slope an-
gle should be reduced because the top layers of soil along the slope
might slip past the lower layers. If the coefficient of static friction be-
tween two such layers is 0.5, what is the least angle fthrough which
the present slope should be reduced to prevent slippage?
θ
90°
Figure 6-51 Problem 67.
68 Engineering a highway curve. If a car goes through a curve too
fast, the car tends to slide out of the curve. For a banked curve with
friction, a frictional force acts on a fast car to oppose the tendency
to slide out of the curve; the force is directed down the bank (in the
direction water would drain). Consider a circular curve of radius
R200 m and bank angle u, where the coefficient of static friction
between tires and pavement is ms. A car (without negative lift) is
driven around the curve as shown in Fig. 6-11. (a) Find an expres-
sion for the car speed vmax that puts the car on the verge of sliding
out. (b) On the same graph, plot vmax versus angle ufor the range 0
to 50, first for ms0.60 (dry pavement) and then for
ms0.050 (wet or icy pavement). In kilometers per hour, evaluate
vmax for a bank angle of u10and for (c) ms0.60 and (d) ms
0.050. (Now you can see why accidents occur in highway curves
when icy conditions are not obvious to drivers, who tend to drive at
normal speeds.)
69 A student, crazed by final exams, uses a force of magnitude
80 N and angle u70to push a 5.0 kg block across the ceiling of
his room (Fig. 6-52). If the coefficient of kinetic friction between the
block and the ceiling is 0.40, what is the magnitude of the block’s
acceleration?
P
:
70 Figure 6-53 shows a conical
pendulum, in which the bob (the
small object at the lower end of the
cord) moves in a horizontal circle at
constant speed. (The cord sweeps
out a cone as the bob rotates.) The
bob has a mass of 0.040 kg, the
string has length L0.90 m and
negligible mass, and the bob follows
a circular path of circumference
0.94 m. What are (a) the tension in
the string and (b) the period of the
motion?
71 An 8.00 kg block of steel is at
rest on a horizontal table. The co-
efficient of static friction between
the block and the table is 0.450. A
force is to be applied to the block.
L
B
ob
Cord
r
Figure 6-53 Problem 70.
θ
Figure 6-54
Problem 73.
77 What is the terminal speed of a 6.00 kg spherical ball that has
a radius of 3.00 cm and a drag coefficient of 1.60? The density of
the air through which the ball falls is 1.20 kg/m3.
78 A student wants to determine the coefficients of static fric-
tion and kinetic friction between a box and a plank. She places
the box on the plank and gradually raises one end of the plank.
When the angle of inclination with the horizontal reaches 30, the
box starts to slip, and it then slides 2.5 m down the plank in 4.0 s
at constant acceleration. What are (a) the coefficient of static
friction and (b) the coefficient of kinetic friction between the box
and the plank?
147
PROBLEMS
79 Block Ain Fig. 6-56 has mass mA4.0 kg, and block Bhas
mass mB2.0 kg.The coefficient of kinetic friction between block B
and the horizontal plane is mk0.50.The inclined plane is frictionless
and at angle u30°.The pulley serves only to change the direction
of the cord connecting the blocks. The cord has negligible mass.
Find (a) the tension in the cord and (b) the magnitude of the accel-
eration of the blocks.
SSM cal change in the road surface because of the temperature de-
crease. By what percentage must the coefficient decrease if the car
is to be in danger of sliding down the street?
86 A sling-thrower puts a stone (0.250 kg) in the sling’s
pouch (0.010 kg) and then begins to make the stone and pouch
move in a vertical circle of radius 0.650 m. The cord between the
pouch and the person’s hand has negligible mass and will break
when the tension in the cord is 33.0 N or more. Suppose the sling-
thrower could gradually increase the speed of the stone. (a) Will
the breaking occur at the lowest point of the circle or at the highest
point? (b) At what speed of the stone will that breaking occur?
87 A car weighing 10.7 kN and traveling at 13.4 m/s without
negative lift attempts to round an unbanked curve with a radius of
61.0 m. (a) What magnitude of the frictional force on the tires is re-
quired to keep the car on its circular path? (b) If the coefficient of
static friction between the tires and the road is 0.350, is the attempt
at taking the curve successful?
88 In Fig. 6-59, block 1 of mass
m12.0 kg and block 2 of mass
m21.0 kg are connected by a
string of negligible mass. Block 2 is
pushed by force of magnitude 20
N and angle u35. The coefficient
of kinetic friction between each block and the horizontal surface is
0.20.What is the tension in the string?
89 A filing cabinet weighing 556 N rests on the floor. The
coefficient of static friction between it and the floor is 0.68, and the
coefficient of kinetic friction is 0.56. In four different attempts to
move it, it is pushed with horizontal forces of magnitudes (a) 222 N,
(b) 334 N, (c) 445 N, and (d) 556 N. For each attempt, calculate the
magnitude of the frictional force on it from the floor. (The cabinet is
initially at rest.) (e) In which of the attempts does the cabinet move?
90 In Fig. 6-60, a block weighing 22 N is held at
rest against a vertical wall by a horizontal force
of magnitude 60 N. The coefficient of static friction
between the wall and the block is 0.55, and the co-
efficient of kinetic friction between them is 0.38. In
six experiments, a second force is applied to the
block and directed parallel to the wall with these
magnitudes and directions: (a) 34 N, up, (b) 12 N,
up, (c) 48 N, up, (d) 62 N, up, (e) 10 N, down, and
(f) 18 N, down. In each experiment, what is the
magnitude of the frictional force on the block? In
which does the block move (g) up the wall and (h) down the wall?
(i) In which is the frictional force directed down the wall?
91 A block slides with constant velocity down an inclined
plane that has slope angle .The block is then projected up the same
plane with an initial speed v0. (a) How far up the plane will it move
before coming to rest? (b) After the block comes to rest, will it slide
down the plane again? Give an argument to back your answer.
92 A circular curve of highway is designed for traffic moving at
60 km/h. Assume the traffic consists of cars without negative lift.
(a) If the radius of the curve is 150 m, what is the correct angle of
banking of the road? (b) If the curve were not banked, what would
be the minimum coefficient of friction between tires and road that
would keep traffic from skidding out of the turn when traveling at
60 km/h?
SSM
P
:
F
:
SSM
F
:
SSM
Frictionless,
massless pulley mB
k
mA
A
B
μ
θ
Figure 6-56 Problem 79.
80 Calculate the magnitude of the drag force on a missile 53 cm
in diameter cruising at 250 m/s at low altitude,where the density of
air is 1.2 kg/m3. Assume C0.75.
81 A bicyclist travels in a circle of radius 25.0 m at a con-
stant speed of 9.00 m/s.The bicycle–rider mass is 85.0 kg. Calculate
the magnitudes of (a) the force of friction on the bicycle from the
road and (b) the net force on the bicycle from the road.
82 In Fig. 6-57, a stuntman drives
a car (without negative lift) over
the top of a hill, the cross section of
which can be approximated by a
circle of radius R250 m. What is
the greatest speed at which he can
drive without the car leaving the road at the top of the hill?
83 You must push a crate across a floor to a docking bay. The
crate weighs 165 N. The coefficient of static friction between crate
and floor is 0.510, and the coefficient of kinetic friction is 0.32.
Your force on the crate is directed horizontally. (a) What magni-
tude of your push puts the crate on the verge of sliding? (b) With
what magnitude must you then push to keep the crate moving at a
constant velocity? (c) If, instead, you then push with the same
magnitude as the answer to (a), what is the magnitude of the
crate’s acceleration?
84 In Fig. 6-58, force is applied
to a crate of mass mon a floor
where the coefficient of static fric-
tion between crate and floor is ms.
Angle uis initially 0but is gradu-
ally increased so that the force vec-
tor rotates clockwise in the figure. During the rotation, the mag-
nitude Fof the force is continuously adjusted so that the crate is
always on the verge of sliding. For ms0.70, (a) plot the ratio
F/mg versus uand (b) determine the angle uinf at which the ratio
approaches an infinite value. (c) Does lubricating the floor in-
crease or decrease uinf, or is the value unchanged? (d) What is uinf
for ms0.60?
85 In the early afternoon, a car is parked on a street that runs
down a steep hill, at an angle of 35.0relative to the horizontal. Just
then the coefficient of static friction between the tires and the
street surface is 0.725. Later, after nightfall, a sleet storm hits the
area, and the coefficient decreases due to both the ice and a chemi-
F
:
SSM
θ
y
x
F
Figure 6-58 Problem 84.
θ
F
m1m2
Figure 6-59 Problem 88.
F
Figure 6-60
Problem 90.
R
Figure 6-57 Problem 82.
148 CHAPTER 6 FORCE AND MOTION—II
93 A 1.5 kg box is initially at rest on a horizontal surface when at
t0 a horizontal force (with tin seconds) is applied
to the box. The acceleration of the box as a function of time tis
given by for 0 t2.8 s and for t
2.8 s. (a) What is the coefficient of static friction between the box
and the surface? (b) What is the coefficient of kinetic friction be-
tween the box and the surface?
94 A child weighing 140 N sits at rest at the top of a playground
slide that makes an angle of 25with the horizontal.The child keeps
from sliding by holding onto the sides of the slide. After letting go
of the sides, the child has a constant acceleration of 0.86 m/s2(down
the slide, of course). (a) What is the coefficient of kinetic friction be-
tween the child and the slide? (b) What maximum and minimum
values for the coefficient of static friction between the child and the
slide are consistent with the information given here?
95 In Fig. 6-61 a fastidious worker
pushes directly along the handle of
a mop with a force . The handle is
at an angle uwith the vertical, and
msand mkare the coefficients of
static and kinetic friction between
the head of the mop and the floor.
Ignore the mass of the handle and
assume that all the mop’s mass mis
in its head. (a) If the mop head
moves along the floor with a con-
stant velocity, then what is F? (b) Show that if uis less than a cer-
tain value u0, then (still directed along the handle) is unable to
move the mop head. Find u0.
96 A child places a picnic basket on the outer rim of a merry-
go-round that has a radius of 4.6 m and revolves once every 30 s.
(a) What is the speed of a point on that rim? (b) What is the lowest
value of the coefficient of static friction between basket and
merry-go-round that allows the basket to stay on the ride?
97 A warehouse worker exerts a constant horizontal force
of magnitude 85 N on a 40 kg box that is initially at rest on the hor-
izontal floor of the warehouse.When the box has moved a distance
of 1.4 m, its speed is 1.0 m/s. What is the coefficient of kinetic fric-
tion between the box and the floor?
98 In Fig. 6-62, a 5.0 kg block is sent sliding up a plane inclined at
u37° while a horizontal force of magnitude 50 N acts on it.
The coefficient of kinetic friction between block and plane is 0.30.
What are the (a) magnitude and (b) direction (up or down the
plane) of the block’s acceleration? The block’s initial speed is 4.0
m/s. (c) How far up the plane does the block go? (d) When it
reaches its highest point, does it remain at rest or slide back down
the plane?
F
:
SSM
F
:
F
:
a
:(1.2t2.4)i
ˆ m/s2
a
:0
F
:(1.8t)i
ˆ N 99 An 11 kg block of steel is at rest on a horizontal table. The
coefficient of static friction between block and table is 0.52. (a)
What is the magnitude of the horizontal force that will put the
block on the verge of moving? (b) What is the magnitude of a
force acting upward 60from the horizontal that will put the
block on the verge of moving? (c) If the force acts downward at
60from the horizontal, how large can its magnitude be without
causing the block to move?
100 A ski that is placed on snow will stick to the snow. However,
when the ski is moved along the snow, the rubbing warms and par-
tially melts the snow, reducing the coefficient of kinetic friction
and promoting sliding.Waxing the ski makes it water repellent and
reduces friction with the resulting layer of water. A magazine
reports that a new type of plastic ski is especially water repellent
and that, on a gentle 200 m slope in the Alps, a skier reduced his
top-to-bottom time from 61 s with standard skis to 42 s with the
new skis. Determine the magnitude of his average acceleration
with (a) the standard skis and (b) the new skis. Assuming a 3.0
slope, compute the coefficient of kinetic friction for (c) the stan-
dard skis and (d) the new skis.
101 Playing near a road construction site, a child falls over a
barrier and down onto a dirt slope that is angled downward at 35
to the horizontal. As the child slides down the slope, he has an
acceleration that has a magnitude of 0.50 m/s2and that is directed
up the slope.What is the coefficient of kinetic friction between the
child and the slope?
102 A 100 N force, directed at an angle uabove a horizontal
floor,is applied to a 25.0 kg chair sitting on the floor. If u0, what
are (a) the horizontal component Fhof the applied force and
(b) the magnitude FNof the normal force of the floor on the chair?
If u30.0, what are (c) Fhand (d) FN? If u60.0, what are (e) Fh
and (f) FN? Now assume that the coefficient of static friction be-
tween chair and floor is 0.420. Does the chair slide or remain at rest
if uis (g) 0, (h) 30.0, and (i) 60.0?
103 A certain string can withstand a maximum tension of 40 N
without breaking.A child ties a 0.37 kg stone to one end and, hold-
ing the other end, whirls the stone in a vertical circle of radius 0.91
m, slowly increasing the speed until the string breaks. (a) Where is
the stone on its path when the string breaks? (b) What is the speed
of the stone as the string breaks?
104 A four-person bobsled (total mass 630 kg) comes
down a straightaway at the start of a bobsled run.The straightaway
is 80.0 m long and is inclined at a constant angle of 10.2with the
horizontal. Assume that the combined effects of friction and air
drag produce on the bobsled a constant force of 62.0 N that acts
parallel to the incline and up the incline. Answer the following
questions to three significant digits. (a) If the speed of the bobsled
at the start of the run is 6.20 m/s, how long does the bobsled take to
come down the straightaway? (b) Suppose the crew is able to re-
duce the effects of friction and air drag to 42.0 N. For the same ini-
tial velocity, how long does the bobsled now take to come down the
straightaway?
105 As a 40 N block slides down a plane that is inclined at 25to
the horizontal, its acceleration is 0.80 m/s2, directed up the plane.
What is the coefficient of kinetic friction between the block and
the plane?
θ
F
Figure 6-61 Problem 95.
F
θ
Figure 6-62 Problem 98.
What Is Physics?
One of the fundamental goals of physics is to investigate something that every-
one talks about: energy. The topic is obviously important. Indeed, our civilization
is based on acquiring and effectively using energy.
For example, everyone knows that any type of motion requires energy:
Flying across the Pacific Ocean requires it. Lifting material to the top floor of an
office building or to an orbiting space station requires it. Throwing a fastball
requires it. We spend a tremendous amount of money to acquire and use energy.
Wars have been started because of energy resources. Wars have been ended
because of a sudden, overpowering use of energy by one side. Everyone knows
many examples of energy and its use, but what does the term energy really mean?
What Is Energy?
The term energy is so broad that a clear definition is difficult to write.Technically,
energy is a scalar quantity associated with the state (or condition) of one or more
objects. However,this definition is too vague to be of help to us now.
A looser definition might at least get us started. Energy is a number that we
associate with a system of one or more objects. If a force changes one of the
objects by, say, making it move, then the energy number changes. After countless
experiments, scientists and engineers realized that if the scheme by which we
assign energy numbers is planned carefully, the numbers can be used to predict the
outcomes of experiments and, even more important, to build machines, such as fly-
ing machines. This success is based on a wonderful property of our universe:
Energy can be transformed from one type to another and transferred from one
object to another, but the total amount is always the same (energy is conserved).
No exception to this principle of energy conservation has ever been found.
Money. Think of the many types of energy as being numbers representing
money in many types of bank accounts. Rules have been made about what such
money numbers mean and how they can be changed. You can transfer money
numbers from one account to another or from one system to another, perhaps
CHAPTER 7
Kinetic Energy and Work
7-1 KINETIC ENERGY
After reading this module, you should be able to . . .
7.01 Apply the relationship between a particle’s kinetic
energy, mass, and speed.
7.02 Identify that kinetic energy is a scalar quantity.
Key Idea
Learning Objectives
149149
●The kinetic energy Kassociated with the motion of a particle of mass mand speed v, where vis well below the speed of light, is
(kinetic energy).K1
2mv2
electronically with nothing material actually moving. However, the total amount
(the total of all the money numbers) can always be accounted for: It is always
conserved. In this chapter we focus on only one type of energy (kinetic energy)
and on only one way in which energy can be transferred (work).
Kinetic Energy
Kinetic energy Kis energy associated with the state of motion of an object. The
faster the object moves, the greater is its kinetic energy. When the object is
stationary, its kinetic energy is zero.
For an object of mass mwhose speed vis well below the speed of light,
(kinetic energy). (7-1)
For example, a 3.0 kg duck flying past us at 2.0 m/s has a kinetic energy of
6.0 kgm2/s2; that is, we associate that number with the duck’s motion.
The SI unit of kinetic energy (and all types of energy) is the joule (J), named
for James Prescott Joule,an English scientist of the 1800s and defined as
1 joule 1J1kgm2/s2. (7-2)
Thus, the flying duck has a kinetic energy of 6.0 J.
K1
2mv2
150 CHAPTER 7 KINETIC ENERGY AND WORK
Sample Problem 7.01 Kinetic energy, train crash
In 1896 in Waco,Texas,William Crush parked two locomotives
at opposite ends of a 6.4-km-long track, fired them up, tied
their throttles open, and then allowed them to crash head-on at
full speed (Fig. 7-1) in front of 30,000 spectators. Hundreds of
people were hurt by flying debris; several were killed.
Assuming each locomotive weighed 1.2 106N and its accel-
eration was a constant 0.26 m/s2, what was the total kinetic en-
ergy of the two locomotives just before the collision?
KEY IDEAS
(1) We need to find the kinetic energy of each locomotive
with Eq. 7-1, but that means we need each locomotive’s
speed just before the collision and its mass. (2) Because we
can assume each locomotive had constant acceleration, we
can use the equations in Table 2-1 to find its speed vjust be-
fore the collision.
Calculations: We choose Eq. 2-16 because we know values
for all the variables except v:
With v00 and xx03.2 103m (half the initial sepa-
ration), this yields
v202(0.26 m/s2)(3.2 103m),
or v40.8 m/s 147 km/h.
v2v0
22a(xx0).
Figure 7-1 The aftermath of an 1896 crash of two locomotives.
Courtesy Library of Congress
We can find the mass of each locomotive by dividing its
given weight by g:
Now, using Eq. 7-1, we find the total kinetic energy of
the two locomotives just before the collision as
(Answer)
This collision was like an exploding bomb.
2.0 108 J.
K2(1
2mv2)(1.22 105 kg)(40.8 m/s)2
m1.2 106 N
9.8 m/s21.22 105 kg.
Additional examples, video, and practice available at WileyPLUS
Work
If you accelerate an object to a greater speed by applying a force to the object,
you increase the kinetic energy of the object. Similarly, if you decel-
erate the object to a lesser speed by applying a force, you decrease the kinetic
energy of the object. We account for these changes in kinetic energy by saying
that your force has transferred energy to the object from yourself or from the
object to yourself. In such a transfer of energy via a force, work Wis said to be
done on the object by the force. More formally, we define work as follows:
K (1
2mv2)
151
7-2 WORK AND KINETIC ENERGY
7-2 WORK AND KINETIC ENERGY
After reading this module, you should be able to . . .
7.03 Apply the relationship between a force (magnitude and
direction) and the work done on a particle by the force
when the particle undergoes a displacement.
7. 0 4 Calculate work by taking a dot product of the force vec-
tor and the displacement vector, in either magnitude-angle
or unit-vector notation.
7.05 If multiple forces act on a particle, calculate the net work
done by them.
7. 0 6 Apply the work–kinetic energy theorem to relate the
work done by a force (or the net work done by multiple
forces) and the resulting change in kinetic energy.
●Work Wis energy transferred to or from an object via a
force acting on the object. Energy transferred to the object
is positive work, and from the object, negative work.
●The work done on a particle by a constant force during
displacement is
(work, constant force),
in which fis the constant angle between the directions of
and .
●Only the component of that is along the displacement
can do work on the object.
d
:
F
:
d
:F
:
WFd cos fF
:d
:
d
:F
:
●When two or more forces act on an object, their net work is
the sum of the individual works done by the forces, which is
also equal to the work that would be done on the object by
the net force of those forces.
●For a particle, a change Kin the kinetic energy equals the
net work Wdone on the particle:
KKfKiW(work–kinetic energy theorem),
in which Kiis the initial kinetic energy of the particle and Kfis
the kinetic energy after the work is done. The equation
rearranged gives us
KfKiW.
F
:
net
Learning Objectives
Key Ideas
Work Wis energy transferred to or from an object by means of a force acting on
the object. Energy transferred to the object is positive work, and energy transferred
from the object is negative work.
“Work,” then, is transferred energy; “doing work” is the act of transferring the
energy. Work has the same units as energy and is a scalar quantity.
The term transfer can be misleading. It does not mean that anything material
flows into or out of the object; that is, the transfer is not like a flow of water.
Rather, it is like the electronic transfer of money between two bank accounts:
The number in one account goes up while the number in the other account goes
down, with nothing material passing between the two accounts.
Note that we are not concerned here with the common meaning of the word
“work,” which implies that any physical or mental labor is work. For example, if
you push hard against a wall, you tire because of the continuously repeated mus-
cle contractions that are required, and you are, in the common sense, working.
However, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol Wonly for work
and shall represent a weight with its equivalent mg.
Work and Kinetic Energy
Finding an Expression for Work
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horizontal xaxis (Fig. 7-2). A constant
force , directed at an angle fto the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton’s second law, written
for components along the xaxis:
Fxmax, (7-3)
where mis the bead’s mass. As the bead moves through a displacement , the
force changes the bead’s velocity from an initial value to some other value .
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write,for components along the xaxis,
(7-4)
Solving this equation for ax, substituting into Eq. 7-3, and rearranging then give us
(7-5)
The first term is the kinetic energy Kfof the bead at the end of the displacement
d, and the second term is the kinetic energy Kiof the bead at the start. Thus, the
left side of Eq. 7-5 tells us the kinetic energy has been changed by the force, and
the right side tells us the change is equal to Fxd. Therefore, the work Wdone on
the bead by the force (the energy transfer due to the force) is
WFxd. (7-6)
If we know values for Fxand d, we can use this equation to calculate the work W.
1
2mv21
2mv0
2Fxd.
v2v0
22axd.
v
:
v
:
0
d
:
F
:
152 CHAPTER 7 KINETIC ENERGY AND WORK
To calculate the work a force does on an object as the object moves through some
displacement, we use only the force component along the object’s displacement.
The force component perpendicular to the displacement does zero work.
Figure 7-2 A constant force directed at
angle fto the displacement of a bead
on a wire accelerates the bead along the
wire, changing the velocity of the bead
from to . A “kinetic energy gauge”
indicates the resulting change in the kinet-
ic energy of the bead, from the value Kito
the value Kf.
In WileyPLUS, this figure is available as
an animation with voiceover.
v
:
v
:
0
d
:
F
:
A
From Fig. 7-2, we see that we can write Fxas Fcos f, where fis the angle
between the directions of the displacement and the force .Thus,
WFd cos f(work done by a constant force). (7-7)
F
:
d
:
xx
Bead
Wire
φ
F
Ki
Kf
v
v0
This component
does no work. Small initial
kinetic energy
Larger final
kinetic energy
This force does positive work
on the bead, increasing speed
and kinetic energy.
This component
does work.
φ
F
φ
F
φ
F
Displacement d
We can use the definition of the scaler (dot) product (Eq. 3-20) to write
(work done by a constant force), (7-8)
where Fis the magnitude of (You may wish to review the discussion of scaler
products in Module 3-3.) Equation 7-8 is especially useful for calculating the
work when and are given in unit-vector notation.
Cautions. There are two restrictions to using Eqs. 7-6 through 7-8 to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Second,
the object must be particle-like. This means that the object must be rigid; all parts
of it must move together, in the same direction. In this chapter we consider only
particle-like objects, such as the bed and its occupant being pushed in Fig.7-3.
Signs for Work. The work done on an object by a force can be either positive
work or negative work. For example,if angle fin Eq. 7-7 is less than 90, then cos fis
positive and thus so is the work. However, if fis greater than 90(up to 180), then
cos fis negative and thus so is the work. (Can you see that the work is zero when
f90?) These results lead to a simple rule. To find the sign of the work done by a
force,consider the force vector component that is parallel to the displacement:
d
:
F
:
F
:
.
WF
:d
:
153
7-2 WORK AND KINETIC ENERGY
Figure 7-3 A contestant in a bed race. We
can approximate the bed and its occupant
as being a particle for the purpose of cal-
culating the work done on them by the
force applied by the contestant.
F
A force does positive work when it has a vector component in the same direction
as the displacement, and it does negative work when it has a vector component in
the opposite direction. It does zero work when it has no such vector component.
Units for Work. Work has the SI unit of the joule, the same as kinetic energy.
However, from Eqs. 7-6 and 7-7 we can see that an equivalent unit is the newton-
meter (Nm). The corresponding unit in the British system is the foot-pound
(ftlb). Extending Eq. 7-2, we have
1J1kgm2/s21Nm0.738 ftlb. (7-9)
Net Work. When two or more forces act on an object, the net work done on
the object is the sum of the works done by the individual forces. We can
calculate the net work in two ways. (1) We can find the work done by each force
and then sum those works. (2) Alternatively, we can first find the net force
of those forces. Then we can use Eq. 7-7, substituting the magnitude Fnet for F
and also the angle between the directions of and for f. Similarly, we can
use Eq. 7-8 with substituted for
Work–Kinetic Energy Theorem
Equation 7-5 relates the change in kinetic energy of the bead (from an initial
to a later ) to the work W(Fxd) done on the bead. ForKf1
2mv2
Ki1
2mv2
0
F
:
.F
:
net
d
:
F
:
net
F
:
net
such particle-like objects, we can generalize that equation. Let Kbe the change
in the kinetic energy of the object, and let Wbe the net work done on it.Then
KKfKiW, (7-10)
which says that
We can also write
KfKiW, (7-11)
which says that
.
kinetic energy after
the net work is done
kinetic energy
before the net work
the net
work done
change in the kinetic
energy of a particle
net work done on
the particle
.
done on the safe by the normal force from the floor?
KEY IDEA
Because these forces are constant in both magnitude and
direction, we can find the work they do with Eq. 7-7.
Calculations: Thus, with mg as the magnitude of the gravi-
tational force, we write
Wgmgd cos 90mgd(0) 0 (Answer)
and WNFNdcos 90FNd(0) 0. (Answer)
We should have known this result. Because these forces are
perpendicular to the displacement of the safe, they do zero
work on the safe and do not transfer any energy to or from it.
(c) The safe is initially stationary. What is its speed vfat the
end of the 8.50 m displacement?
KEY IDEA
The speed of the safe changes because its kinetic energy is
changed when energy is transferred to it by and .F
:
2
F
:
1
F
:
N
These statements are known traditionally as the work– kinetic energy theorem
for particles. They hold for both positive and negative work: If the net work done
on a particle is positive, then the particle’s kinetic energy increases by the amount
of the work. If the net work done is negative, then the particle’s kinetic energy
decreases by the amount of the work.
For example, if the kinetic energy of a particle is initially 5 J and there is a
net transfer of 2 J to the particle (positive net work), the final kinetic energy is
7 J. If, instead, there is a net transfer of 2 J from the particle (negative net work),
the final kinetic energy is 3 J.
154 CHAPTER 7 KINETIC ENERGY AND WORK
Checkpoint 1
A particle moves along an xaxis.Does the kinetic energy of the particle increase,de-
crease, or remain the same if the particle’s velocity changes (a) from 3 m/s to 2 m/s
and (b) from 2 m/s to 2 m/s? (c) In each situation, is the work done on the particle
positive, negative, or zero?
Sample Problem 7.02 Work done by two constant forces, industrial spies
8.50 m. The push of spy 001 is 12.0 N at an angle of 30.0
downward from the horizontal; the pull of spy 002 isF
:
2
F
:
1
Figure 7-4 (a) Two spies move a floor safe through a displacement
.(b) A free-body diagram for the safe.
d
:
(a)
Safe
(b)
40.0°
30.0°
Spy 001 Spy 002
Fg
FN
F1
F2
d
Only force components
parallel to the displacement
do work.
(b) During the displacement, what is the work Wgdone on the
safe by the gravitational force and what is the work WN
F
:
g
Figure 7-4ashows two industrial spies sliding an initially
stationary 225 kg floor safe a displacement of magnituded
:
10.0 N at 40.0above the horizontal.The magnitudes and di-
rections of these forces do not change as the safe moves, and
the floor and safe make frictionless contact.
(a) What is the net work done on the safe by forces and
during the displacement ?
KEY IDEAS
(1) The net work Wdone on the safe by the two forces is the
sum of the works they do individually. (2) Because we can
treat the safe as a particle and the forces are constant in
both magnitude and direction, we can use either Eq. 7-7
(WFd cos f) or Eq. 7-8 to calculate those
works. Let’s choose Eq.7-7.
Calculations: From Eq. 7-7 and the free-body diagram for
the safe in Fig.7-4b, the work done by is
W1F1dcos f1(12.0 N)(8.50 m)(cos 30.0)
88.33 J,
and the work done by is
W2F2dcos f2(10.0 N)(8.50 m)(cos 40.0)
65.11 J.
Thus, the net work Wis
WW1W288.33 J 65.11 J
153.4 J 153 J. (Answer)
During the 8.50 m displacement, therefore, the spies transfer
153 J of energy to the kinetic energy of the safe.
F
:
2
F
:
1
(WF
:d
:)
d
:F
:
2
F
:
1
155
7-3 WORK DONE BY THE GRAVITATIONAL FORCE
Calculations: We relate the speed to the work done by
combining Eqs. 7-10 (the work–kinetic energy theorem) and
7-1 (the definition of kinetic energy):
The initial speed viis zero, and we now know that the work
WKfKi1
2mvf
21
2mvi
2.
done is 153.4 J. Solving for vfand then substituting known
data, we find that
(Answer)1.17 m/s.
vfA2W
mA2(153.4 J)
225 kg
Sample Problem 7.03 Work done by a constant force in unit-vector notation
During a storm, a crate of crepe is sliding across a slick,
oily parking lot through a displacement
while a steady wind pushes against the crate with a force
. The situation and coordinate
axes are shown in Fig.7-5.
(a) How much work does this force do on the crate during
the displacement?
KEY IDEA
Because we can treat the crate as a particle and because the
wind force is constant (“steady”) in both magnitude and direc-
tion during the displacement, we can use either Eq. 7-7 (W
Fd cos f) or Eq. 7-8 to calculate the work. Since
we know and in unit-vector notation, we choose Eq. 7-8.
Calculations: We write
Of the possible unit-vector dot products, only i
ˆi
ˆ,ˆ
jˆ
j, and
ˆ
kˆ
k are nonzero (see Appendix E). Here we obtain
W(2.0 N)(3.0 m)i
ˆi
ˆ(6.0 N)(3.0 m)ˆ
ji
ˆ
(6.0 J)(1) 06.0 J. (Answer)
WF
:d
:[(2.0 N)i
ˆ(6.0 N)j
ˆ][(3.0 m)i
ˆ].
d
:
F
:(WF
:d
:)
(2.0 N)i
ˆ(6.0 N)j
ˆ
F
:
d
:(3.0 m)i
ˆ
Figure 7-5 Force slows a
crate during displacement .
d
:
F
:
y
x
F
d
The parallel force component does
negative work, slowing the crate.
Thus, the force does a negative 6.0 J of work on the crate,trans-
ferring 6.0 J of energy from the kinetic energy of the crate.
(b) If the crate has a kinetic energy of 10 J at the beginning
of displacement , what is its kinetic energy at the end of ?
KEY IDEA
Because the force does negative work on the crate, it re-
duces the crate’s kinetic energy.
Calculation: Using the work–kinetic energy theorem in
the form of Eq. 7-11, we have
KfKiW10 J (6.0 J) 4.0 J. (Answer)
Less kinetic energy means that the crate has been slowed.
d
:
d
:
Additional examples, video, and practice available at WileyPLUS
7-3 WORK DONE BY THE GRAVITATIONAL FORCE
Learning Objectives
7. 0 8 Apply the work–kinetic energy theorem to situations
where an object is lifted or lowered.
●The work Wgdone by the gravitational force on a
particle-like object of mass mas the object moves through a
displacement is given by
Wgmgd cos f,
in which fis the angle between and .
●The work Wadone by an applied force as a particle-like
object is either lifted or lowered is related to the work Wg
d
:
F
:
g
d
:
F
:
gdone by the gravitational force and the change Kin the
object’s kinetic energy by
KKfKiWaWg.
If KfKi, then the equation reduces to
WaWg,
which tells us that the applied force transfers as much energy
to the object as the gravitational force transfers from it.
After reading this module, you should be able to . . .
7. 0 7 Calculate the work done by the gravitational force
when an object is lifted or lowered.
Key Ideas
156 CHAPTER 7 KINETIC ENERGY AND WORK
Work Done by the Gravitational Force
We next examine the work done on an object by the gravitational force acting on
it. Figure 7-6 shows a particle-like tomato of mass mthat is thrown upward with
initial speed v0and thus with initial kinetic energy . As the tomatoKi1
2mv2
0
Figure 7-6 Because the gravitational force
acts on it, a particle-like tomato of mass m
thrown upward slows from velocity to
velocity during displacement . A kinetic
energy gauge indicates the resulting change
in the kinetic energy of the tomato, from
to .Kf(1
2mv2)Ki(1
2mv2
0)
d
:
v
:
v
:
0
F
:
g
Kf
Ki
Fg
Fg
Fg
v0
v
d
The force does negative
work, decreasing speed
and kinetic energy.
rises, it is slowed by a gravitational force ; that is, the tomato’s kinetic energyF
:
g
decreases because does work on the tomato as it rises. Because we can treatF
:
g
Figure 7-7 (a) An applied force lifts an
object. The object’s displacement makes
an angle f180with the gravitational
force on the object. The applied force
does positive work on the object. (b) An
applied force lowers an object. The dis-
placement of the object makes an angle
fwith the gravitational force . The
applied force does negative work on the
object.
F
:
g
0
d
:F
:
F
:
g
d
:
F
:
(Fig. 7-7a), then f180and the work done by the applied force equals mgd.
the tomato as a particle, we can use Eq. 7-7 (WFd cos f) to express the work
done during a displacement . For the force magnitude F, we use mg as the mag-
nitude of F
:
g.Thus, the work Wgdone by the gravitational force F
:
gis
Wgmgd cos f(work done by gravitational force). (7-12)
For a rising object, force F
:
gis directed opposite the displacement , as indi-
cated in Fig.7-6.Thus, f180and
Wgmgd cos 180mgd(1) mgd. (7-13)
The minus sign tells us that during the object’s rise, the gravitational force acting
on the object transfers energy in the amount mgd from the kinetic energy of the
object.This is consistent with the slowing of the object as it rises.
After the object has reached its maximum height and is falling back down,
the angle fbetween force and displacement is zero.Thus,
Wgmgd cos 0mgd(1) mgd. (7-14)
The plus sign tells us that the gravitational force now transfers energy in the amount
mgd to the kinetic energy of the falling object (it speeds up,of course).
Work Done in Lifting and Lowering an Object
Now suppose we lift a particle-like object by applying a vertical force to it.
During the upward displacement, our applied force does positive work Waon the
object while the gravitational force does negative work Wgon it. Our applied
force tends to transfer energy to the object while the gravitational force tends to
transfer energy from it. By Eq. 7-10, the change Kin the kinetic energy of the
object due to these two energy transfers is
KKfKiWaWg, (7-15)
in which Kfis the kinetic energy at the end of the displacement and Kiis that at
the start of the displacement. This equation also applies if we lower the object,
but then the gravitational force tends to transfer energy to the object while our
force tends to transfer energy from it.
If an object is stationary before and after a lift (as when you lift a book from
the floor to a shelf), then Kfand Kiare both zero, and Eq. 7-15 reduces to
WaWg0
or WaWg. (7-16)
Note that we get the same result if Kfand Kiare not zero but are still equal.
Either way, the result means that the work done by the applied force is the nega-
tive of the work done by the gravitational force; that is, the applied force transfers
the same amount of energy to the object as the gravitational force transfers from
the object. Using Eq. 7-12, we can rewrite Eq. 7-16 as
Wamgd cos f(work done in lifting and lowering; KfKi), (7-17)
with fbeing the angle between and . If the displacement is vertically upwardd
:
F
:
g
F
:
d
:
F
:
g
d
:
d
:
(a)
Fg
F
d
Object
Does
positive
work
Upward
displacement
Does
negative
work
(b)
Fg
F
d
Object
Does
positive
work
Downward
displacement
Does
negative
work
157
7-3 WORK DONE BY THE GRAVITATIONAL FORCE
If the displacement is vertically downward (Fig. 7-7b), then f0and the work
done by the applied force equals mgd.
Equations 7-16 and 7-17 apply to any situation in which an object is lifted or
lowered, with the object stationary before and after the lift.They are independent
of the magnitude of the force used. For example, if you lift a mug from the floor
to over your head, your force on the mug varies considerably during the lift. Still,
because the mug is stationary before and after the lift, the work your force does
on the mug is given by Eqs. 7-16 and 7-17, where, in Eq. 7-17, mg is the weight of
the mug and dis the distance you lift it.
The angle fbetween the displacement and this force com-
ponent is 180. So we can apply Eq. 7-7 to write
Sample Problem 7.04 Work in pulling a sleigh up a snowy slope
In this problem an object is pulled along a ramp but the ob-
ject starts and ends at rest and thus has no overall change in
its kinetic energy (that is important). Figure 7-8ashows the
situation.A rope pulls a 200 kg sleigh (which you may know)
up a slope at incline angle u30, through distance d20 m.
The sleigh and its contents have a total mass of 200 kg. The
snowy slope is so slippery that we take it to be frictionless.
How much work is done by each force acting on the sleigh?
KEY IDEAS
(1) During the motion, the forces are constant in magnitude
and direction and thus we can calculate the work done by
each with Eq. 7-7 (WFd cos f) in which fis the angle be-
tween the force and the displacement. We reach the same
result with Eq. 7-8 (W) in which we take a dot prod-
uct of the force vector and displacement vector. (2) We can
relate the net work done by the forces to the change in
kinetic energy (or lack of a change, as here) with the
work–kinetic energy theorem of Eq. 7-10 (KW).
Calculations: The first thing to do with most physics prob-
lems involving forces is to draw a free-body diagram to organ-
ize our thoughts. For the sleigh, Fig.7-8bis our free-body dia-
gram, showing the gravitational force , the force from theT
:
F
:
g
d
:
F
:
rope,and the normal force from the slope.
Work WNby the normal force. Let’s start with this easy cal-
culation.The normal force is perpendicular to the slope and
thus also to the sleigh’s displacement.Thus the normal force
does not affect the sleigh’s motion and does zero work. To
be more formal, we can apply Eq. 7-7 to write
WNFNdcos 90 0. (Answer)
Work Wgby the gravitational force. We can find the work
done by the gravitational force in either of two ways (you
pick the more appealing way). From an earlier discussion
about ramps (Sample Problem 5.04 and Fig. 5-15), we know
that the component of the gravitational force along the
slope has magnitude mg sin uand is directed down the
slope.Thus the magnitude is
Fgx mg sin u(200 kg)(9.8 m/s2) sin 30
980 N.
FN
:
WgFgxdcos 180(980 N)(20 m)(1)
Figure 7-8 (a) A sleigh is pulled up a snowy slope. (b) The free-
body diagram for the sleigh.
θ
d
F
N
T
F
g
mg cosu
mg sinu
(b)
(a)
u
Does
positive work
Does negative work
x
1.96 104J. (Answer)
The negative result means that the gravitational force re-
moves energy from the sleigh.
The second (equivalent) way to get this result is to use
the full gravitational force instead of a component. TheF
:
g
angle between and is 120 (add the incline angle 30d
:
F
:
g
to 90). So, Eq. 7-7 gives us
WgFgdcos 120mgd cos 120
(200 kg)(9.8 m/s2)(20 m) cos 120
1.96 104J. (Answer)
Work WTby the rope’s force. We have two ways of calculat-
ing this work. The quickest way is to use the work–kinetic en-
ergy theorem of Eq. 7-10 (KW), where the net work W
done by the forces is WNWgWTand the change Kin the
kinetic energy is just zero (because the initial and final kinetic
energies are the same—namely, zero). So,Eq.7-10 gives us
0WNWgWT01.96 104JWT
and WT1.96 l04J. (Answer)
158 CHAPTER 7 KINETIC ENERGY AND WORK
Sample Problem 7.05 Work done on an accelerating elevator cab
An elevator cab of mass m500 kg is descending with speed
vi4.0 m/s when its supporting cable begins to slip, allowing
it to fall with constant acceleration (Fig.7-9a).
(a) During the fall through a distance d12 m, what is the
work Wgdone on the cab by the gravitational force ?
KEY IDEA
We can treat the cab as a particle and thus use Eq. 7-12
(Wgmgd cos f) to find the work Wg.
Calculation: From Fig. 7-9b, we see that the angle between
the directions of F
:
gand the cab’s displacement is 0.So,
Wgmgd cos 0(500 kg)(9.8 m/s2)(12 m)(1)
5.88 104J59 kJ. (Answer)
(b) During the 12 m fall, what is the work WTdone on the
cab by the upward pull of the elevator cable?
KEY IDEA
We can calculate work WTwith Eq. 7-7 (WFd cos f) by
first writing Fnet,ymayfor the components in Fig.7-9b.
Calculations: We get
TFgma. (7-18)
Solving for T, substituting mg for Fg, and then substituting
the result in Eq. 7-7, we obtain
WTTd cos fm(ag)dcos f. (7-19)
Next, substituting g/5 for the (downward) acceleration a
and then 180for the angle fbetween the directions of
forces and , we find
(Answer)4.70 104 J 47 kJ.
4
5 (500 kg)(9.8 m/s2)(12 m) cos 180
WTm
g
5g
d cos
4
5mgd cos
mg
:
T
:
T
:
d
:
F
:
g
a
:g
:/5
Figure 7-9 An elevator
cab, descending with
speed vi, suddenly
begins to accelerate
downward. (a) It
moves through a dis-
placement with
constant acceleration
(b) A free-
body diagram for the
cab, displacement
included.
a
:g
:/5.
d
:
Caution: Note that WTis not simply the negative of Wgbe-
cause the cab accelerates during the fall. Thus, Eq. 7-16
(which assumes that the initial and final kinetic energies are
equal) does not apply here.
(c) What is the net work Wdone on the cab during the fall?
Calculation: The net work is the sum of the works done by
the forces acting on the cab:
WWgWT5.88 104J4.70 104J
1.18 104J12 kJ. (Answer)
(d) What is the cab’s kinetic energy at the end of the 12 m fall?
KEY IDEA
The kinetic energy changes because of the net work done on
the cab, according to Eq. 7-11 (KfKiW).
Calculation: From Eq. 7-1, we write the initial kinetic
energy as .We then write Eq. 7-11 as
(Answer)1.58 104 J 16 kJ.
1
2(500 kg)(4.0 m/s)21.18 104 J
KfKiW1
2mvi
2W
Ki1
2mvi
2
Additional examples, video, and practice available at WileyPLUS
Instead of doing this, we can apply Newton’s second law for
motion along the xaxis to find the magnitude FTof the rope’s
force. Assuming that the acceleration along the slope is zero
(except for the brief starting and stopping), we can write
Fnet,xmax,
FTmg sin 30m(0),
to find
FTmg sin 30.
This is the magnitude. Because the force and the displace-
ment are both up the slope, the angle between those two
vectors is zero. So, we can now write Eq. 7-7 to find the work
done by the rope’s force:
WTFTdcos 0(mg sin 30)dcos 0
(200 kg)(9.8 m/s2)(sin 30)(20 m) cos 0
1.96 104J. (Answer)
Elevator
cable
Cab
(b)(a)
a
d
Fg
T
y
Does
negative
work
Does
positive
work
159
7-4 WORK DONE BY A SPRING FORCE
Work Done by a Spring Force
We next want to examine the work done on a particle-like object by a particular
type of variable force —namely, a spring force, the force from a spring. Many
forces in nature have the same mathematical form as the spring force. Thus, by
examining this one force, you can gain an understanding of many others.
The Spring Force
Figure 7-10ashows a spring in its relaxed state —that is, neither compressed nor
extended. One end is fixed, and a particle-like object—a block, say—is attached
to the other, free end. If we stretch the spring by pulling the block to the right as
in Fig. 7-10b, the spring pulls on the block toward the left. (Because a spring
force acts to restore the relaxed state, it is sometimes said to be a restoring force.)
If we compress the spring by pushing the block to the left as in Fig. 7-10c, the
spring now pushes on the block toward the right.
To a good approximation for many springs, the force from a spring is pro-
portional to the displacement of the free end from its position when the spring
is in the relaxed state.The spring force is given by
(Hooke’s law), (7-20)
which is known as Hooke’s law after Robert Hooke, an English scientist of the
late 1600s. The minus sign in Eq. 7-20 indicates that the direction of the spring
force is always opposite the direction of the displacement of the spring’s free end.
The constant kis called the spring constant (or force constant) and is a measure
of the stiffness of the spring.The larger kis, the stiffer the spring; that is, the larger
kis, the stronger the spring’s pull or push for a given displacement.The SI unit for
kis the newton per meter.
In Fig. 7-10 an xaxis has been placed parallel to the length of the spring, with
the origin (x0) at the position of the free end when the spring is in its relaxed
F
:
skd
:
d
:F
:
s
7-4 WORK DONE BY A SPRING FORCE
Learning Objectives
position of the object or by using the known generic result
of that integration.
7.12 Calculate work by graphically integrating on a graph of
force versus position of the object.
7.13 Apply the work–kinetic energy theorem to situations in
which an object is moved by a spring force.
●The force from a spring is
(Hooke’s law),
where is the displacement of the spring’s free end from
its position when the spring is in its relaxed state (neither
compressed nor extended), and kis the spring constant
(a measure of the spring’s stiffness). If an xaxis lies along the
spring, with the origin at the location of the spring’s free end
when the spring is in its relaxed state, we can write
Fxkx (Hooke’s law).
d
:
F
:
skd
:
F
:
s●A spring force is thus a variable force: It varies with the
displacement of the spring’s free end.
●If an object is attached to the spring’s free end, the work Ws
done on the object by the spring force when the object is
moved from an initial position xito a final position xfis
If xi0and xfx, then the equation becomes
Ws
1
2kx2.
Ws1
2kxi
21
2kxf
2.
After reading this module, you should be able to . . .
7. 0 9 Apply the relationship (Hooke’s law) between the force
on an object due to a spring, the stretch or compression
of the spring, and the spring constant of the spring.
7.10 Identify that a spring force is a variable force.
7.11 Calculate the work done on an object by a spring force
by integrating the force from the initial position to the final
Key Ideas
Figure 7-10 (a) A spring in its relaxed state.
The origin of an xaxis has been placed at
the end of the spring that is attached to a
block. (b) The block is displaced by , and
the spring is stretched by a positive amount
x. Note the restoring force exerted by
the spring. (c) The spring is compressed by
a negative amount x.Again, note the
restoring force.
F
:
s
d
:
Block
attached
to spring
x
0
x
0
x
x
0
x
x= 0
Fx= 0
xpositive
Fxnegative
xnegative
Fxpositive
(a)
(b)
(c)
d
d
Fs
Fs
160 CHAPTER 7 KINETIC ENERGY AND WORK
state. For this common arrangement, we can write Eq. 7-20 as
Fxkx (Hooke’s law), (7-21)
where we have changed the subscript. If xis positive (the spring is stretched
toward the right on the xaxis), then Fxis negative (it is a pull toward the left). If
xis negative (the spring is compressed toward the left), then Fxis positive (it is a
push toward the right). Note that a spring force is a variable force because it is a
function of x, the position of the free end.Thus Fxcan be symbolized as F(x).Also
note that Hooke’s law is a linear relationship between Fxand x.
The Work Done by a Spring Force
To find the work done by the spring force as the block in Fig. 7-10amoves, let us
make two simplifying assumptions about the spring. (1) It is massless; that is, its
mass is negligible relative to the block’s mass. (2) It is an ideal spring; that is, it
obeys Hooke’s law exactly. Let us also assume that the contact between the block
and the floor is frictionless and that the block is particle-like.
We give the block a rightward jerk to get it moving and then leave it alone.
As the block moves rightward, the spring force Fxdoes work on the block,
decreasing the kinetic energy and slowing the block. However, we cannot find this
work by using Eq. 7-7 (WFd cos f) because there is no one value of Fto plug
into that equation—the value of Fincreases as the block stretches the spring.
There is a neat way around this problem. (1) We break up the block’s dis-
placement into tiny segments that are so small that we can neglect the variation
in Fin each segment. (2) Then in each segment, the force has (approximately) a
single value and thus we can use Eq. 7-7 to find the work in that segment. (3)
Then we add up the work results for all the segments to get the total work. Well,
that is our intent, but we don’t really want to spend the next several days adding
up a great many results and, besides, they would be only approximations. Instead,
let’s make the segments infinitesimal so that the error in each work result goes to
zero. And then let’s add up all the results by integration instead of by hand.
Through the ease of calculus, we can do all this in minutes instead of days.
Let the block’s initial position be xiand its later position be xf. Then divide
the distance between those two positions into many segments, each of tiny length
x. Label these segments, starting from xi, as segments 1, 2, and so on. As the
block moves through a segment, the spring force hardly varies because the seg-
ment is so short that xhardly varies. Thus, we can approximate the force magni-
tude as being constant within the segment. Label these magnitudes as Fx1in
segment 1, Fx2in segment 2, and so on.
With the force now constant in each segment, we can find the work done
within each segment by using Eq. 7-7. Here f180, and so cos f1. Then
the work done is Fx1xin segment 1, Fx2xin segment 2, and so on. The net
work Wsdone by the spring,from xito xf, is the sum of all these works:
(7-22)
where jlabels the segments. In the limit as xgoes to zero, Eq. 7-22 becomes
(7-23)
From Eq. 7-21, the force magnitude Fxis kx.Thus, substitution leads to
(7-24)(1
2k)[x2]xi
xf(1
2k)(xf
2xi
2).
Wsxf
xi
kx dx kxf
xi
xdx
Wsxf
xi
Fxdx.
Ws
Fxj x,
161
Caution: If the block is not stationary before and after the displacement, then this
statement is not true.
Work Wsis positive if the block ends up closer to the relaxed position (x0) than
it was initially. It is negative if the block ends up farther away from x0. It is zero
if the block ends up at the same distance from x0.
If xi0 and if we call the final position x, then Eq. 7-25 becomes
(work by a spring force). (7-26)
The Work Done by an Applied Force
Now suppose that we displace the block along the xaxis while continuing to apply a
force to it. During the displacement, our applied force does work Waon the blockF
:
a
Ws
1
2kx2
7-4 WORK DONE BY A SPRING FORCE
If a block that is attached to a spring is stationary before and after a displacement,
then the work done on it by the applied force displacing it is the negative of the
work done on it by the spring force.
Checkpoint 2
For three situations,the initial and final positions, respectively, along the xaxis for the
block in Fig.7-10 are (a) 3 cm, 2 cm;(b) 2 cm, 3 cm; and (c) 2 cm, 2 cm. In each sit-
uation, is the work done by the spring force on the block positive, negative,or zero?
Sample Problem 7.06 Work done by a spring to change kinetic energy
When a spring does work on an object, we cannot find the
work by simply multiplying the spring force by the object’s
displacement. The reason is that there is no one value for
the force—it changes. However, we can split the displace-
ment up into an infinite number of tiny parts and then ap-
proximate the force in each as being constant. Integration
sums the work done in all those parts. Here we use the
generic result of the integration.
In Fig. 7-11, a cumin canister of mass m0.40 kg slides
across a horizontal frictionless counter with speed v0.50 m/s.
Multiplied out, this yields
(work by a spring force). (7-25)
This work Wsdone by the spring force can have a positive or negative value,
depending on whether the net transfer of energy is to or from the block as the
block moves from xito xf.Caution: The final position xfappears in the second
term on the right side of Eq. 7-25.Therefore, Eq. 7-25 tells us:
Ws1
2kxi
21
2kxf
2
Figure 7-11 A canister moves toward a spring.
k
m
Frictionless
First touchStop
v
d
The spring force does
negative work, decreasing
speed and kinetic energy.
while the spring force does work Ws. By Eq. 7-10, the change Kin the kinetic en-
ergy of the block due to these two energy transfers is
KKfKiWaWs, (7-27)
in which Kfis the kinetic energy at the end of the displacement and Kiis that at
the start of the displacement. If the block is stationary before and after the dis-
placement, then Kfand Kiare both zero and Eq.7-27 reduces to
WaWs. (7-28)
162 CHAPTER 7 KINETIC ENERGY AND WORK
Work Done by a General Variable Force
One-Dimensional Analysis
Let us return to the situation of Fig. 7-2 but now consider the force to be in the
positive direction of the xaxis and the force magnitude to vary with position x.
Thus, as the bead (particle) moves, the magnitude F(x) of the force doing work on
it changes. Only the magnitude of this variable force changes, not its direction,
and the magnitude at any position does not change with time.
7-5 WORK DONE BY A GENERAL VARIABLE FORCE
After reading this module, you should be able to . . .
7.14 Given a variable force as a function of position, calculate
the work done by it on an object by integrating the function
from the initial to the final position of the object, in one or
more dimensions.
7.15 Given a graph of force versus position, calculate the
work done by graphically integrating from the initial
position to the final position of the object.
7.16 Convert a graph of acceleration versus position to a
graph of force versus position.
7. 17 Apply the work–kinetic energy theorem to situations
where an object is moved by a variable force.
Learning Objectives
●When the force on a particle-like object depends on
the position of the object, the work done by on the ob-
ject while the object moves from an initial position riwith
coordinates (xi,yi,zi)to a final position rfwith coordinates
(xf,yf,zf)must be found by integrating the force. If we as-
sume that component Fxmay depend on xbut not on yor
z, component Fymay depend on ybut not on xor z, and
component Fzmay depend on zbut not on xor y, then the
F
:
F
:work is
●If has only an xcomponent, then this reduces to
Wxf
xi
F(x)dx.
F
:
Wxf
xi
Fxdx yf
yi
Fydy zf
zi
Fzdz.
Key Ideas
Additional examples, video, and practice available at WileyPLUS
It then runs into and compresses a spring of spring constant
k750 N/m. When the canister is momentarily stopped by
the spring,by what distance dis the spring compressed?
KEY IDEAS
1. The work Wsdone on the canister by the spring force is
related to the requested distance dby Eq. 7-26 (Ws
, with dreplacing x.
2. The work Wsis also related to the kinetic energy of the
canister by Eq. 7-10 (KfKiW).
3. The canister’s kinetic energy has an initial value of K
and a value of zero when the canister is momen-
tarily at rest.
1
2mv2
1
2kx2)
Calculations: Putting the first two of these ideas together,
we write the work–kinetic energy theorem for the canister as
Substituting according to the third key idea gives us this
expression:
Simplifying, solving for d, and substituting known data then
give us
(Answer)1.2 102 m 1.2 cm.
dvAm
k(0.50 m/s) A0.40 kg
750 N/m
01
2mv2
1
2kd2.
KfKi
1
2kd2.
163
7-5 WORK DONE BY A GENERAL VARIABLE FORCE
Figure 7-12 (a) A one-dimensional force
plotted against the displacement xof
a particle on which it acts. The particle
moves from xito xf.(b) Same as (a) but
with the area under the curve divided into
narrow strips. (c) Same as (b) but with the
area divided into narrower strips. (d) The
limiting case. The work done by the force
is given by Eq. 7-32 and is represented by
the shaded area between the curve and
the xaxis and between xiand xf.
F
:
(x)
Figure 7-12ashows a plot of such a one-dimensional variable force. We want
an expression for the work done on the particle by this force as the particle
moves from an initial point xito a final point xf. However, we cannot use Eq. 7-7
(WFd cos f) because it applies only for a constant force . Here, again, we
shall use calculus. We divide the area under the curve of Fig. 7-12ainto a number
of narrow strips of width x(Fig. 7-12b).We choose xsmall enough to permit us
to take the force F(x) as being reasonably constant over that interval.We let Fj,avg
be the average value of F(x) within the jth interval.Then in Fig. 7-12b,Fj,avg is the
height of the jth strip.
With Fj,avg considered constant, the increment (small amount) of work
Wjdone by the force in the jth interval is now approximately given by Eq.
7-7 and is
WjFj,avg x. (7-29)
In Fig. 7-12b,Wjis then equal to the area of the jth rectangular, shaded strip.
To approximate the total work Wdone by the force as the particle moves
from xito xf, we add the areas of all the strips between xiand xfin Fig. 7-12b:
Wx. (7-30)
Equation 7-30 is an approximation because the broken “skyline” formed by the tops
of the rectangular strips in Fig.7-12bonly approximates the actual curve of F(x).
We can make the approximation better by reducing the strip width xand
using more strips (Fig. 7-12c). In the limit, we let the strip width approach
zero; the number of strips then becomes infinitely large and we have, as an ex-
act result,
(7-31)
This limit is exactly what we mean by the integral of the function F(x) between
the limits xiand xf.Thus, Eq. 7-31 becomes
(work: variable force). (7-32)
If we know the function F(x), we can substitute it into Eq. 7-32, introduce the
proper limits of integration, carry out the integration, and thus find the work.
(Appendix E contains a list of common integrals.) Geometrically, the work is
equal to the area between the F(x) curve and the xaxis, between the limits xiand
xf(shaded in Fig.7-12d).
Three-Dimensional Analysis
Consider now a particle that is acted on by a three-dimensional force
FxFyFz, (7-33)
in which the components Fx,Fy, and Fzcan depend on the position of the particle;
that is, they can be functions of that position. However, we make three simplifica-
tions: Fxmay depend on xbut not on yor z,Fymay depend on ybut not on xor z,
and Fzmay depend on zbut not on xor y.Now let the particle move through an in-
cremental displacement
dx dy dz . (7-34)
The increment of work dW done on the particle by during the displacement
is, by Eq. 7-8,
(7-35)dW F
:dr
:Fxdx Fydy Fzdz.
dr
:
F
:
k
ˆ
j
ˆ
i
ˆ
dr
:
k
ˆ
j
ˆ
i
ˆ
F
:
Wxf
xi
F(x)dx
Wlim
x:0
Fj,avg x.
Fj,avg
Wj
F
:
F(x)
x
xixf
0
(a)
Work is equal to the
area under the curve.
F(x)
x
xixf
F
j
, avg
Δ
x
0
(b)
ΔW
j
We can approximate that area
with the area of these strips.
F(x)
x
xixf
0
Δ
x
(c)
We can do better with
more, narrower strips.
F(x)
x
xixf
0
W
(d)
For the best, take the limit of
strip widths going to zero.
164 CHAPTER 7 KINETIC ENERGY AND WORK
Sample Problem 7.07 Work calculated by graphical integration
In Fig. 7-13b, an 8.0 kg block slides along a frictionless floor
as a force acts on it, starting at x10 and ending at x36.5 m.
As the block moves, the magnitude and direction of the
force varies according to the graph shown in Fig. 7-13a.For
The work Wdone by while the particle moves from an initial position rihaving
coordinates (xi,yi,zi) to a final position rfhaving coordinates (xf,yf,zf) is then
(7-36)
If has only an xcomponent, then the yand zterms in Eq. 7-36 are zero and the
equation reduces to Eq. 7-32.
Work–Kinetic Energy Theorem with a Variable Force
Equation 7-32 gives the work done by a variable force on a particle in a one-
dimensional situation. Let us now make certain that the work is equal to the
change in kinetic energy, as the work–kinetic energy theorem states.
Consider a particle of mass m, moving along an xaxis and acted on by a
net force F(x) that is directed along that axis. The work done on the particle
by this force as the particle moves from position xito position xfis given by
Eq. 7-32 as
(7-37)
in which we use Newton’s second law to replace F(x) with ma. We can write the
quantity ma dx in Eq. 7-37 as
(7-38)
From the chain rule of calculus, we have
(7-39)
and Eq. 7-38 becomes
(7-40)
Substituting Eq. 7-40 into Eq. 7-37 yields
(7-41)
Note that when we change the variable from xto vwe are required to express the
limits on the integral in terms of the new variable. Note also that because the
mass mis a constant, we are able to move it outside the integral.
Recognizing the terms on the right side of Eq. 7-41 as kinetic energies allows
us to write this equation as
WKfKiK,
which is the work–kinetic energy theorem.
1
2mvf
21
2mvi
2.
Wvf
vi
mv dv mvf
vi
vdv
ma dx mdv
dx vdxmv dv.
dv
dt dv
dx
dx
dt dv
dx v,
ma dx mdv
dt dx.
Wxf
xi
F(x)dx xf
xi
ma dx,
F
:
Wrf
ri
dW xf
xi
Fxdx yf
yi
Fydy zf
zi
Fzdz.
F
:
example, from x0 to x1 m, the force is positive (in
the positive direction of the xaxis) and increases in mag-
nitude from 0 to 40 N. And from x4 m to x5 m, the
force is negative and increases in magnitude from 0 to 20 N.
165
7-5 WORK DONE BY A GENERAL VARIABLE FORCE
Additional examples, video, and practice available at WileyPLUS
2
0
⫺20
40
46
2046
x (m)
x (m)
F (N)
(a)
(b)
v
1
v
2
v
3
F F
Figure 7-13 (a) A graph indicating the magnitude and direction of a
variable force that acts on a block as it moves along an xaxis on
a floor, (b) The location of the block at several times.
Again using the definition of kinetic energy, we find
and then
(Answer)
This is the block’s greatest speed because from x4.0 m to
x6.5 m the force is negative, meaning that it opposes the
block’s motion, doing negative work on the block and thus
decreasing the kinetic energy and speed. In that range, the
area between the plot and the xaxis is
This means that the work done by the force in that range is
35 J. At x4.0, the block has K400 J. At x6.5 m, the
work–kinetic energy theorem tells us that its kinetic energy is
Again using the definition of kinetic energy, we find
and then
(Answer)
The block is still moving in the positive direction of the
xaxis, a bit faster than initially.
v39.55 m/s 9.6 m/s.
365 J 1
2(8.0 kg)v2
3,
K31
2mv2
3,
400 J 35 J 365 J.
K3K2W
35 J.
1
2(20 N)(1 m) (20 N)(1 m) 1
2(20 N)(0.5 m) 35 Nm
v210 m/s.
400 J 1
2(8.0 kg)v2
2,
K21
2mv2
2,
(Note that this latter value is displayed as 20 N.) The
block’s kinetic energy at x1is K1280 J. What is the
block’s speed at x10, x24.0 m, and x36.5 m?
KEY IDEAS
(1) At any point, we can relate the speed of the block to its
kinetic energy with Eq. 7-1 (2) We can relate
the kinetic energy Kfat a later point to the initial kinetic Ki
and the work Wdone on the block by using the work–
kinetic energy theorem of Eq. 7-10 (KfKiW). (3) We
can calculate the work Wdone by a variable force F(x) by
integrating the force versus position x. Equation 7-32 tells
us that
We don’t have a function F(x) to carry out the integration,
but we do have a graph of F(x) where we can integrate by
finding the area between the plotted line and the xaxis.
Where the plot is above the axis, the work (which is equal to
the area) is positive. Where it is below the axis, the work is
negative.
Calculations: The requested speed at x0 is easy because
we already know the kinetic energy. So, we just plug the
kinetic energy into the formula for kinetic energy:
and then
(Answer)
As the block moves from x0 to x4.0 m, the plot in
Figure 7-13ais above the xaxis, which means that positive
work is being done on the block.We split the area under the
plot into a triangle at the left, a rectangle in the center, and a
triangle at the right.Their total area is
This means that between x0 and x4.0 m, the force
does 120 J of work on the block, increasing the kinetic en-
ergy and speed of the block. So, when the block reaches
x4.0 m, the work–kinetic energy theorem tells us that
the kinetic energy is
280 J 120 J 400 J.
K2K1W
120 J.
1
2(40 N)(1 m) (40 N)(2 m) 1
2(40 N)(1 m) 120 Nm
v18.37 m/s 8.4 m/s.
280 J 1
2(8.0 kg)v2
1,
K11
2mv2
1,
Wxf
xi
F(x)dx.
(K1
2mv2).
166 CHAPTER 7 KINETIC ENERGY AND WORK
KEY IDEA
The force is a variable force because its xcomponent de-
pends on the value of x. Thus, we cannot use Eqs. 7-7 and 7-8
to find the work done. Instead, we must use Eq. 7-36 to inte-
grate the force.
Calculation: We set up two integrals, one along each axis:
(Answer)
The positive result means that energy is transferred to the
particle by force . Thus, the kinetic energy of the particle
increases and, because , its speed must also
increase. If the work had come out negative, the kinetic
energy and speed would have decreased.
K1
2mv2
F
:
7.0 J.
3[1
3x3]2
34[y]3
0[3323]4[0 3]
W3
23x2dx 0
3 4 dy 33
2x2dx 40
3dy
Sample Problem 7.08 Work, two-dimensional integration
When the force on an object depends on the position of the
object, we cannot find the work done by it on the object by
simply multiplying the force by the displacement. The rea-
son is that there is no one value for the force—it changes.
So, we must find the work in tiny little displacements and
then add up all the work results.We effectively say,“Yes, the
force varies over any given tiny little displacement, but the
variation is so small we can approximate the force as being
constant during the displacement.” Sure, it is not precise, but
if we make the displacements infinitesimal, then our error
becomes infinitesimal and the result becomes precise. But,
to add an infinite number of work contributions by hand
would take us forever, longer than a semester. So, we add
them up via an integration, which allows us to do all this in
minutes (much less than a semester).
Force (3x2N) (4 N) , with xin meters, acts on a
particle, changing only the kinetic energy of the particle.
How much work is done on the particle as it moves from co-
ordinates (2 m, 3 m) to (3 m, 0 m)? Does the speed of the
particle increase, decrease, or remain the same?
j
ˆ
i
ˆ
F
:
Additional examples, video, and practice available at WileyPLUS
7-6 POWER
Learning Objectives
7.20 Determine the instantaneous power by taking a dot
product of the force vector and an object’s velocity vector,
in magnitude-angle and unit-vector notations.
●The power due to a force is the rate at which that force
does work on an object.
●If the force does work Wduring a time interval t, the aver-
age power due to the force over that time interval is
Pavg W
t.
●Instantaneous power is the instantaneous rate of doing work:
●For a force at an angle fto the direction of travel of the
instantaneous velocity , the instantaneous power is
.PFv cos
F
:v
:
v
:
F
:
PdW
dt .
After reading this module, you should be able to . . .
7.18 Apply the relationship between average power, the
work done by a force, and the time interval in which that
work is done.
7.19 Given the work as a function of time, find the instanta-
neous power.
Key Ideas
Power
The time rate at which work is done by a force is said to be the power due to the
force. If a force does an amount of work Win an amount of time t, the average
power due to the force during that time interval is
(average power). (7-42)Pavg W
t
167
The instantaneous power Pis the instantaneous time rate of doing work, which
we can write as
(instantaneous power). (7-43)
Suppose we know the work W(t) done by a force as a function of time. Then to
get the instantaneous power Pat, say, time t3.0 s during the work, we would
first take the time derivative of W(t) and then evaluate the result for t3.0 s.
The SI unit of power is the joule per second.This unit is used so often that it
has a special name, the watt (W), after James Watt, who greatly improved the
rate at which steam engines could do work. In the British system, the unit of
power is the foot-pound per second. Often the horsepower is used. These are
related by
1 watt 1W1 J/s 0.738 ftlb/s (7-44)
and 1 horsepower 1hp550 ft lb/s 746 W. (7-45)
Inspection of Eq. 7-42 shows that work can be expressed as power multiplied
by time, as in the common unit kilowatt-hour.Thus,
1 kilowatt-hour 1kWh(103W)(3600 s)
3.60 106J3.60 MJ. (7-46)
Perhaps because they appear on our utility bills, the watt and the kilowatt-hour
have become identified as electrical units. They can be used equally well as units
for other examples of power and energy. Thus, if you pick up a book from the
floor and put it on a tabletop, you are free to report the work that you have done
as, say, 4 106kWh (or more conveniently as 4 mWh).
We can also express the rate at which a force does work on a particle (or
particle-like object) in terms of that force and the particle’s velocity. For a par-
ticle that is moving along a straight line (say, an xaxis) and is acted on by a
constant force directed at some angle fto that line, Eq. 7-43 becomes
or PFv cos f. (7-47)
Reorganizing the right side of Eq. 7-47 as the dot product we may also write
the equation as
(instantaneous power). (7-48)
For example, the truck in Fig. 7-14 exerts a force on the trailing load, which
has velocity at some instant. The instantaneous power due to is the rate at
which does work on the load at that instant and is given by Eqs. 7-47 and 7-48.
Saying that this power is “the power of the truck” is often acceptable, but keep in
mind what is meant: Power is the rate at which the applied force does work.
F
:F
:
v
:
F
:
PF
:v
:
F
:v
:,
PdW
dt F cos f
dx
dt F cos f
dx
dt
,
F
:
PdW
dt
7-6 POWER
Figure 7-14 The power due to the truck’s
applied force on the trailing load is the
rate at which that force does work on the
load.
© Reglain/ZUMA
Checkpoint 3
A block moves with uniform circular motion because a cord tied to the block is an-
chored at the center of a circle. Is the power due to the force on the block from the
cord positive, negative, or zero?
168 CHAPTER 7 KINETIC ENERGY AND WORK
Calculation: We use Eq. 7-47 for each force.For force , at
angle f1180to velocity , we have
P1F1vcos f1(2.0 N)(3.0 m/s) cos 180
6.0 W. (Answer)
This negative result tells us that force is transferring en-
ergy from the box at the rate of 6.0 J/s.
For force , at angle f260to velocity , we have
P2F2vcos f2(4.0 N)(3.0 m/s) cos 60
6.0 W. (Answer)
This positive result tells us that force is transferring en-
ergy to the box at the rate of 6.0 J/s.
The net power is the sum of the individual powers
(complete with their algebraic signs):
Pnet P1P2
6.0 W 6.0 W 0, (Answer)
which tells us that the net rate of transfer of energy to
or from the box is zero.Thus, the kinetic energy
of the box is not changing, and so the speed of the box will
remain at 3.0 m/s.With neither the forces and nor the
velocity changing, we see from Eq. 7-48 that P1and P2are
constant and thus so is Pnet.
v
:
F
:
2
F
:
1
(K1
2mv2)
F
:
2
v
:
F
:
2
F
:
1
v
:
F
:
1
Sample Problem 7.09 Power, force, and velocity
Here we calculate an instantaneous work—that is, the rate at
which work is being done at any given instant rather than av-
eraged over a time interval. Figure 7-15 shows constant forces
and acting on a box as the box slides rightward across a
frictionless floor.Force is horizontal, with magnitude 2.0 N;F
:
1
F
:
2
F
:
1
Additional examples, video, and practice available at WileyPLUS
Figure 7-15 Two forces and act on a box that slides
rightward across a frictionless floor. The velocity of the box is .v
:
F
:
2
F
:
1
60°
Frictionless F1
F2
v
Negative power.
(This force is
removing energy.)
Positive power.
(This force is
supplying energy.)
Kinetic Energy The kinetic energy Kassociated with the mo-
tion of a particle of mass mand speed v, where vis well below the
speed of light, is
(kinetic energy). (7-1)
Work Work Wis energy transferred to or from an object via a
force acting on the object. Energy transferred to the object is posi-
tive work, and from the object, negative work.
Work Done by a Constant Force The work done on a par-
ticle by a constant force during displacement is
(work, constant force), (7-7, 7-8)
in which fis the constant angle between the directions of and .
Only the component of that is along the displacement can do
work on the object. When two or more forces act on an object,
their net work is the sum of the individual works done by the
forces, which is also equal to the work that would be done on the
object by the net force of those forces.
Work and Kinetic Energy For a particle, a change Kin the
kinetic energy equals the net work Wdone on the particle:
KKfKiW(work–kinetic energy theorem), (7-10)
F
:
net
d
:
F
:d
:
F
:
WFd cos
F
:d
:
d
:
F
:
K1
2mv2
Review & Summary
in which Kiis the initial kinetic energy of the particle and Kfis the ki-
netic energy after the work is done. Equation 7-10 rearranged gives us
KfKiW. (7-11)
Work Done by the Gravitational Force The work Wg
done by the gravitational force on a particle-like object of mass
mas the object moves through a displacement is given by
Wgmgd cos f, (7-12)
in which fis the angle between and .
Work Done in Lifting and Lowering an Object The work
Wadone by an applied force as a particle-like object is either lifted
or lowered is related to the work Wgdone by the gravitational
force and the change Kin the object’s kinetic energy by
KKfKiWaWg. (7-15)
If KfKi, then Eq. 7-15 reduces to
WaWg, (7-16)
which tells us that the applied force transfers as much energy to the
object as the gravitational force transfers from it.
d
:
F
:
g
d
:
F
:
g
force is angled upward by 60to the floor and has magni-
tude 4.0 N.The speed vof the box at a certain instant is 3.0 m/s.
What is the power due to each force acting on the box at that
instant, and what is the net power? Is the net power changing
at that instant?
KEY IDEA
We want an instantaneous power, not an average power
over a time period.Also, we know the box’s velocity (rather
than the work done on it).
F
:
2
169
QUESTIONS
F
x
F
1
–F
1
x
1
x
(a)
F
x
F
1
–F
1
x
1
x
(b)
F
x
F
1
–F
1
x
1
x
(c)
F
x
F
1
–F
1
x
1
x
(d)
Figure 7-18
Question 5.
Spring Force The force from a spring is
(Hooke’s law), (7-20)
where is the displacement of the spring’s free end from its posi-
tion when the spring is in its relaxed state (neither compressed nor
extended), and kis the spring constant (a measure of the spring’s
stiffness). If an xaxis lies along the spring, with the origin at the lo-
cation of the spring’s free end when the spring is in its relaxed
state, Eq. 7-20 can be written as
Fxkx (Hooke’s law). (7-21)
A spring force is thus a variable force: It varies with the
displacement of the spring’s free end.
Work Done by a Spring Force If an object is attached to
the spring’s free end, the work Wsdone on the object by the spring
force when the object is moved from an initial position xito a final
position xfis
(7-25)
If xi0 and xfx, then Eq. 7-25 becomes
(7-26)
Work Done by a Variable Force When the force on a particle-
like object depends on the position of the object, the work done by
on the object while the object moves from an initial position riwith co-
ordinates (xi,yi,zi) to a final position rfwith coordinates (xf,yf,zf)
F
:
F
:
Ws
1
2kx2.
Ws1
2kxi
21
2kxf
2.
d
:
F
:
skd
:
F
:
smust be found by integrating the force. If we assume that component
Fxmay depend on xbut not on yor z, component Fymay depend on y
but not on xor z, and component Fzmay depend on zbut not on xor
y,then the work is
(7-36)
If has only an xcomponent, then Eq. 7-36 reduces to
(7-32)
Power The power due to a force is the rate at which that force
does work on an object. If the force does work Wduring a time inter-
val t,the average power due to the force over that time interval is
(7-42)
Instantaneous power is the instantaneous rate of doing work:
(7-43)
For a force at an angle fto the direction of travel of the instan-
taneous velocity , the instantaneous power is
. (7-47, 7-48)PFv cos
F
:v
:
v
:
F
:
PdW
dt .
Pavg W
t.
Wxf
xi
F(x)dx.
F
:
Wxf
xi
Fxdx yf
yi
Fydy zf
zi
Fzdz.
Questions
1Rank the following velocities according to the kinetic energy a
particle will have with each velocity, greatest first: (a) ,
(b) , (c) , (d) , (e) ,v
:5i
ˆ
3i
ˆ4j
ˆ
v
:v
:3i
ˆ4j
ˆ
v
:4i
ˆ3j
ˆv
:4i
ˆ3j
ˆ
F2
F1
(a) (b)
3
2
1
K
t
Figure 7-16 Question 2.
3Is positive or negative work done by a constant force on a par-
ticle during a straight-line displacement if (a) the angle between
and is 30;(b) the angle is 100; (c) and ?
4In three situations, a briefly applied horizontal force changes the
velocity of a hockey puck that slides over frictionless ice. The over-
head views of Fig. 7-17 indicate, for each situation, the puck’s initial
speed vi,its final speed vf, and the directions of the corresponding ve-
locity vectors. Rank the situations according to the work done on the
puck by the applied force, most positive first and most negative last.
d
:4i
ˆ
F
:2i
ˆ3j
ˆ
d
:F
:
d
:F
:
Figure 7-17 Question 4.
and (f) v5 m/s at 30to the horizontal.
2Figure 7-16ashows two horizontal forces that act on a block
that is sliding to the right across a frictionless floor. Figure 7-16b
shows three plots of the block’s kinetic energy Kversus time t.
Which of the plots best corresponds to the following three situ-
ations: (a) F1F2, (b) F1F2, (c) F1F2?
5The graphs in Fig. 7-18 give the xcomponent Fxof a force act-
ing on a particle moving along an xaxis. Rank them according to
the work done by the force on the particle from x0 to xx1,
from most positive work first to most negative work last.
(a)(b)(c)
y
v
f
= 5 m/s
v
i
= 6 m/s x
y
v
f
= 3 m/s
v
i
= 4 m/s x
yv
f
= 4 m/s
v
i
= 2 m/s
x
170 CHAPTER 7 KINETIC ENERGY AND WORK
6Figure 7-19 gives the xcom-
ponent Fxof a force that can act
on a particle. If the particle be-
gins at rest at x0, what is its
coordinate when it has (a) its
greatest kinetic energy, (b) its
greatest speed, and (c) zero
speed? (d) What is the particle’s
direction of travel after it
reaches x6m?
7In Fig. 7-20, a greased pig has a choice of three frictionless slides
along which to slide to the ground. Rank the slides according to how
much work the gravitational force does on the pig during the descent,
greatest first.
8Figure 7-21ashows four situations in which a horizontal force acts
on the same block, which is initially at rest. The force magnitudes are
F2F42F12F3. The horizontal component vxof the block’s ve-
locity is shown in Fig. 7-21bfor the four situations. (a) Which plot in
Fig. 7-21bbest corresponds to which force in Fig. 7-21a? (b) Which
1 2 3 4 5 6 7 8
x (m)
F
2
F1
Fx
–F1
–F
2
Figure 7-19 Question 6.
(a) (b) (c)
Figure 7-20
Question 7.
F1F2F4
F3
x
(a)
(b)
vx
t
D
C
B
A
(c)
K
t
HG
F
E
Figure 7-21 Question 8.
K
K
K
K
K
K
K
K
t
t
t
t
t
t
t
t
(a) (b) (c) (d)
(e) (
f
) (
g
) (h)
Figure 7-22 Question 10.
11 In three situations, a single force acts on a moving particle.
Here are the velocities (at that instant) and the forces:
(1) (2)
(3) . Rank
the situations according to the rate at which energy is being trans-
ferred, greatest transfer to the particle ranked first, greatest trans-
fer from the particle ranked last.
12 Figure 7-23 shows three arrangements of a block attached to
identical springs that are in their relaxed state when the block is
centered as shown. Rank the arrangements according to the mag-
nitude of the net force on the block, largest first, when the block is
displaced by distance d(a) to the right and (b) to the left. Rank the
arrangements according to the work done on the block by the
spring forces, greatest first, when the block is displaced by d(c) to
the right and (d) to the left.
F
:(2i
ˆ6j
ˆ) Nv
:(3i
ˆj
ˆ) m/s,F
:(2j
ˆ7k
ˆ) N;
v
:(2i
ˆ3j
ˆ) m/s,F
:(6i
ˆ20j
ˆ) N;v
:(4i
ˆ) m/s,
(1) (2) (3)
Figure 7-23 Question 12.
plot in Fig. 7-21c(for kinetic energy Kversus time t) best corre-
sponds to which plot in Fig.7-21b?
9Spring Ais stiffer than spring B(kAkB). The spring force of
which spring does more work if the springs are compressed (a) the
same distance and (b) by the same applied force?
10 A glob of slime is launched or dropped from the edge of a
cliff. Which of the graphs in Fig. 7-22 could possibly show how the
kinetic energy of the glob changes during its flight?
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Module 7-1 Kinetic Energy
•1 A proton (mass m1.67 1027 kg) is being acceler-
ated along a straight line at 3.6 1015 m/s2in a machine. If the pro-
ton has an initial speed of 2.4 107m/s and travels 3.5 cm, what
then is (a) its speed and (b) the increase in its kinetic energy?
SSM
•2 If a Saturn V rocket with an Apollo spacecraft attached had a
combined mass of 2.9 105kg and reached a speed of 11.2 km/s,
how much kinetic energy would it then have?
•3 On August 10, 1972, a large meteorite skipped across the
atmosphere above the western United States and western Canada,
171
PROBLEMS
much like a stone skipped across water.The accompanying fireball
was so bright that it could be seen in the daytime sky and was
brighter than the usual meteorite trail. The meteorite’s mass was
about 4 106kg; its speed was about 15 km/s. Had it entered the
atmosphere vertically, it would have hit Earth’s surface with about
the same speed. (a) Calculate the meteorite’s loss of kinetic energy
(in joules) that would have been associated with the vertical impact.
(b) Express the energy as a multiple of the explosive energy of
1 megaton of TNT, which is 4.2 1015 J. (c) The energy associated
with the atomic bomb explosion over Hiroshima was equivalent to
13 kilotons of TNT. To how many Hiroshima bombs would the me-
teorite impact have been equivalent?
•4 An explosion at ground level leaves a crater with a diam-
eter that is proportional to the energy of the explosion raised to
the power; an explosion of 1 megaton of TNT leaves a crater
with a 1 km diameter. Below Lake Huron in Michigan there ap-
pears to be an ancient impact crater with a 50 km diameter. What
was the kinetic energy associated with that impact, in terms of
(a) megatons of TNT (1 megaton yields 4.2 1015 J) and
(b) Hiroshima bomb equivalents (13 kilotons of TNT each)?
(Ancient meteorite or comet impacts may have significantly
altered the climate, killing off the dinosaurs and other life-forms.)
••5 A father racing his son has half the kinetic energy of the son,
who has half the mass of the father.The father speeds up by 1.0 m/s
and then has the same kinetic energy as the son.What are the origi-
nal speeds of (a) the father and (b) the son?
••6 A bead with mass 1.8 102kg is moving along a wire in
the positive direction of an xaxis. Beginning at time t0, when
the bead passes through x0 with speed 12 m/s, a constant force
acts on the bead. Figure 7-24 indicates the bead’s position at
these four times: t00, t11.0 s, t22.0 s, and t33.0 s. The
bead momentarily stops at t3.0 s. What is the kinetic energy of
the bead at t10 s?
1
3
0 5 10 15 20
t0t1t2t3
x (m)
Figure 7-24 Problem 6.
0 0.2 0.4 0.6 0.8
t = 0 0.5 s 1.0 s 1.5 s 2.0 s
x (m)
Figure 7-25 Problem 7.
ity of 4.0 m/s in the positive xdirection and some time later has a
velocity of 6.0 m/s in the positive ydirection. How much work is
done on the canister by the 5.0 N force during this time?
•10 A coin slides over a frictionless plane and across an xy
coordinate system from the origin to a point with xy coordinates
(3.0 m, 4.0 m) while a constant force acts on it.The force has mag-
nitude 2.0 N and is directed at a counterclockwise angle of 100
from the positive direction of the xaxis. How much work is done
by the force on the coin during the displacement?
••11 A 12.0 N force with a fixed orientation does work on a
particle as the particle moves through the three-dimensional dis-
placement m. What is the angle be-
tween the force and the displacement if the change in the particle’s
kinetic energy is (a) 30.0 J and (b) 30.0 J?
••12 A can of bolts and nuts is
pushed 2.00 m along an xaxis by a
broom along the greasy (friction-
less) floor of a car repair shop in a
version of shuffleboard. Figure 7-26
gives the work Wdone on the can
by the constant horizontal force
from the broom, versus the can’s po-
sition x.The scale of the figure’s ver-
tical axis is set by Ws6.0 J. (a)
What is the magnitude of that
force? (b) If the can had an initial kinetic energy of 3.00 J, moving
in the positive direction of the xaxis, what is its kinetic energy at
the end of the 2.00 m?
••13 A luge and its rider, with a total mass of 85 kg, emerge from a
downhill track onto a horizontal straight track with an initial speed
of 37 m/s. If a force slows them to a stop at a constant rate of 2.0
m/s2, (a) what magnitude Fis required for the force, (b) what dis-
tance ddo they travel while slowing, and (c) what work Wis done
on them by the force? What are (d) F, (e) d, and (f) Wif they, in-
stead, slow at 4.0 m/s2?
••14 Figure 7-27 shows an over-
head view of three horizontal forces
acting on a cargo canister that was
initially stationary but now moves
across a frictionless floor. The force
magnitudes are F13.00 N, F2
4.00 N, and F310.0 N, and the indi-
cated angles are u250.0and u3
35.0. What is the net work done on
the canister by the three forces dur-
ing the first 4.00 m of displacement?
••15 Figure 7-28 shows three
forces applied to a trunk that moves
leftward by 3.00 m over a friction-
less floor. The force magnitudes are
F15.00 N, F29.00 N, and F3
3.00 N, and the indicated angle is u
60.0. During the displacement,
(a) what is the net work done on the
trunk by the three forces and (b)
does the kinetic energy of the trunk
increase or decrease?
••16 An 8.0 kg object is moving in the positive direction
of an x
axis.When it passes through x0, a constant force directed
d
:(2.00i
ˆ4.00j
ˆ3.00k
ˆ)
W (J)
W
s
0 1
x (m)
2
Figure 7-26 Problem 12.
Module 7-2 Work and Kinetic Energy
•7 A 3.0 kg body is at rest on a frictionless horizontal air track
when a constant horizontal force acting in the positive direction of
an xaxis along the track is applied to the body.A stroboscopic graph
of the position of the body as it slides to the right is shown in Fig. 7-
25.The force is applied to the body at t0, and the graph records
the position of the body at 0.50 s intervals. How much work is done
on the body by the applied force between t0 and t2.0 s?F
:
F
:
F
:
•8 A ice block floating in a river is pushed through a displacement
along a straight embankment by rushing wa-
ter,which exerts a force on the block. How
much work does the force do on the block during the displacement?
•9 The only force acting on a 2.0 kg canister that is moving in an
xy plane has a magnitude of 5.0 N.The canister initially has a veloc-
F
:(210 N)i
ˆ(150 N)j
ˆ
d
:(15 m)i
ˆ(12 m)j
ˆ
F
1
F
2
F
3
y
x
2
θ
3
θ
Figure 7-27 Problem 14.
θ
F1
F3
F2
Figure 7-28 Problem 15.
172 CHAPTER 7 KINETIC ENERGY AND WORK
of magnitude 20.0 N is applied to a
3.00 kg psychology book as the book
slides a distance d0.500 m up a fric-
tionless ramp at angle u30.0. (a)
During the displacement, what is the net
work done on the book by , the gravi-
tational force on the book, and the nor-
mal force on the book? (b) If the book
has zero kinetic energy at the start of the
displacement, what is its speed at the end of the displacement?
•••25 In Fig.7-34, a 0.250 kg block of cheese lies on
the floor of a 900 kg elevator cab that is being pulled
upward by a cable through distance d12.40 m and
then through distance d210.5 m. (a) Through d1,if
the normal force on the block from the floor has con-
stant magnitude FN3.00 N, how much work is done
on the cab by the force from the cable? (b) Through d2,
if the work done on the cab by the (constant) force
from the cable is 92.61 kJ, what is the magnitude of FN?
Module 7-4 Work Done by a Spring Force
•26 In Fig. 7-10, we must apply a force of magnitude 80 N to hold the
block stationary at x2.0 cm. From that position, we then slowly
move the block so that our force does 4.0 J of work on the
spring–block system; the block is then again stationary. What is the
block’s position? (Hint: There are two answers.)
•27 A spring and block are in the arrangement of Fig. 7-10.When the
block is pulled out to x4.0 cm, we must apply a force of magnitude
360 N to hold it there.We pull the block to x11 cm and then release
it. How much work does the spring do on the block as the block
moves from xi5.0 cm to (a) x 3.0 cm, (b) x3.0 cm,
(c) x5.0 cm,and (d) x9.0 cm?
•28 During spring semester at MIT, residents of the parallel build-
ings of the East Campus dorms battle one another with large cata-
pults that are made with surgical hose mounted on a window frame.
A balloon filled with dyed water is placed in a pouch attached to the
hose, which is then stretched through the width of the room.Assume
that the stretching of the hose obeys Hooke’s law with a spring con-
stant of 100 N/m. If the hose is stretched by 5.00 m and then released,
how much work does the force from the hose do on the balloon in
the pouch by the time the hose reaches its relaxed length?
••29 In the arrangement of Fig. 7-10, we gradually pull the block
from x0 to x3.0 cm, where it is stationary. Figure 7-35 gives
F
:
a
F
:
a
through vertical distance h0.150 m?
••24 In Fig. 7-33, a horizontal force
along the axis begins to act on it.
Figure 7-29 gives its kinetic energy
Kversus position xas it moves
from x0 to x5.0 m; K030.0
J. The force continues to act. What
is vwhen the object moves back
through x3.0 m?
Module 7-3 Work Done by
the Gravitational Force
•17 A helicopter lifts a 72 kg astronaut 15 m verti-
cally from the ocean by means of a cable. The acceleration of the
astronaut is g/10. How much work is done on the astronaut by
(a) the force from the helicopter and (b) the gravitational force on
her? Just before she reaches the helicopter,what are her (c) kinetic
energy and (d) speed?
•18 (a) In 1975 the roof of Montreal’s Velodrome, with
a weight of 360 kN, was lifted by 10 cm so that it could be centered.
How much work was done on the roof by the forces making the
lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one
end of a car that had fallen onto her son when a jack failed. If her
panic lift effectively raised 4000 N (about of the car’s weight) by
5.0 cm, how much work did her force do on the car?
••19 In Fig. 7-30, a block of ice
slides down a frictionless ramp at angle
50while an ice worker pulls on
the block (via a rope) with a force
that has a magnitude of 50 N and is di-
rected up the ramp. As the block slides
through distance d0.50 m along the
ramp, its kinetic energy increases by 80
J. How much greater would its kinetic
energy have been if the rope had not
been attached to the block?
••20 A block is sent up a frictionless
ramp along which an xaxis extends up-
ward. Figure 7-31 gives the kinetic en-
ergy of the block as a function of posi-
tion x; the scale of the figure’s vertical
axis is set by Ks40.0 J. If the block’s
initial speed is 4.00 m/s, what is the nor-
mal force on the block?
••21 A cord is used to vertically
lower an initially stationary block of
mass Mat a constant downward acceleration of g/4.When the block
has fallen a distance d, find (a) the work done by the cord’s force on
the block, (b) the work done by the gravitational force on the block,
(c) the kinetic energy of the block, and (d) the speed of the block.
••22 A cave rescue team lifts an injured spelunker directly upward
and out of a sinkhole by means of a motor-driven cable. The lift is
performed in three stages, each requiring a vertical distance of 10.0
m: (a) the initially stationary spelunker is accelerated to a speed of
5.00 m/s; (b) he is then lifted at the con-
stant speed of 5.00 m/s; (c) finally he is
decelerated to zero speed. How much
work is done on the 80.0 kg rescuee by
the force lifting him during each stage?
••23 In Fig. 7-32, a constant force of
magnitude 82.0 N is applied to a 3.00
kg shoe box at angle 53.0, causing
F
:
a
SSM
F
:
r
1
4
WWWSSM
the box to move up a frictionless ramp at constant speed. How
θ
d
F
r
Figure 7-30 Problem 19.
K
s
0 1
x (m)
K (J)
2
Figure 7-31 Problem 20.
Figure 7-29 Problem 16.
x (m)
K
0
05
K (J)
much work is done on the box by when the box has movedF
:
a
φ
Fa
Figure 7-32 Problem 23.
θ
Psychology
Psychology
d
Fa
Figure 7-33 Problem 24.
Figure 7-34
Problem 25.
0
x (cm)
123
W (J)
Ws
Figure 7-35 Problem 29.
173
PROBLEMS
a (m/s
2
)
a
s
0
x(m)
4 6 8 20
Figure 7-38 Problem 34.
Figure 7-40 Problem 37.
h
x
T
x2x1
y
Figure 7-41 Problem 42.
the work that our force does on the block.The scale of the figure’s
vertical axis is set by Ws1.0 J. We then pull the block out to x
5.0 cm and release it from rest. How much work does the spring
do on the block when the block moves from xi5.0 cm to
(a) x4.0 cm, (b) x2.0 cm, and (c) x5.0 cm?
••30 In Fig. 7-10a, a block of mass
mlies on a horizontal frictionless
surface and is attached to one end
of a horizontal spring (spring con-
stant k) whose other end is fixed.
The block is initially at rest at the
position where the spring is
unstretched (x0) when a con-
stant horizontal force in the positive direction of the xaxis is ap-
plied to it.A plot of the resulting kinetic energy of the block versus
its position xis shown in Fig. 7-36. The scale of the figure’s vertical
axis is set by Ks4.0 J. (a) What is the magnitude of ? (b) What
is the value of k?
••31 The only force acting on a 2.0 kg body as it
moves along a positive xaxis has an xcomponent Fx6xN,
with xin meters.The velocity at x3.0 m is 8.0 m/s. (a) What is the
velocity of the body at x4.0 m? (b) At what positive value of x
will the body have a velocity of 5.0 m/s?
••32 Figure 7-37 gives spring force
Fxversus position xfor the
spring–block arrangement of Fig. 7-
10. The scale is set by Fs160.0 N.
We release the block at x12 cm.
How much work does the spring do
on the block when the block moves
from xi8.0 cm to (a) x5.0
cm, (b) x5.0 cm, (c) x8.0
cm,and (d) x10.0 cm?
•••33 The block in Fig. 7-10alies on a horizontal frictionless
surface, and the spring constant is 50 N/m. Initially, the spring is at
its relaxed length and the block is stationary at position x0.
Then an applied force with a constant magnitude of 3.0 N pulls the
block in the positive direction of the xaxis, stretching the spring
until the block stops.When that stopping point is reached, what are
(a) the position of the block, (b) the work that has been done on
the block by the applied force, and (c) the work that has been done
on the block by the spring force? During the block’s displacement,
what are (d) the block’s position when its kinetic energy is maxi-
mum and (e) the value of that maximum kinetic energy?
Module 7-5 Work Done by a General Variable Force
•34 A 10 kg brick moves along an xaxis. Its acceleration as a
function of its position is shown in Fig. 7-38.The scale of the figure’s
vertical axis is set by as20.0 m/s2. What is the net work per-
formed on the brick by the force causing the acceleration as the
brick moves from x0 to x8.0 m?
ILW
WWWSSM
F
:
F
:
•35 The force on a particle is directed along an xaxis
and given by FF
0(x/x01). Find the work done by the force in
moving the particle from x0 to x2x0by (a) plotting F(x) and
measuring the work from the graph and (b) integrating F(x).
•36 A 5.0 kg block moves in a
straight line on a horizontal friction-
less surface under the influence of a
force that varies with position as
shown in Fig. 7-39.The scale of the fig-
ure’s vertical axis is set by Fs10.0 N.
How much work is done by the force
as the block moves from the origin
to x8.0 m?
••37 Figure 7-40 gives the accel-
eration of a 2.00 kg particle as an applied force moves it from restF
:
a
WWWSSM
K (J)
Ks
00 0.5 1
x (m)
1.5 2
Figure 7-36 Problem 30.
x (cm)
–2 –1 0
Fs
F
x
–Fs
1 2
Figure 7-37 Problem 32.
Force (N)
Fs
0
–Fs
4
Position (m)
82
Figure 7-39 Problem 36.
••38 A 1.5 kg block is initially at rest on a horizontal frictionless
surface when a horizontal force along an xaxis is applied to the block.
The force is given by , where xis in meters and
the initial position of the block is x0. (a) What is the kinetic energy
of the block as it passes through x2.0 m? (b) What is the maximum
kinetic energy of the block between x0 and x2.0 m?
••39 A force acts on a particle as the parti-
cle moves along an xaxis, with in newtons, xin meters, and ca
constant.At x0, the particle’s kinetic energy is 20.0 J; at x3.00 m,
it is 11.0 J. Find c.
••40 A can of sardines is made to move along an xaxis from
x0.25 m to x1.25 m by a force with a magnitude given by
Fexp(4x2),with xin meters and Fin newtons.(Here exp is the ex-
ponential function.) How much work is done on the can by the force?
••41 A single force acts on a 3.0 kg particle-like object whose posi-
tion is given by x3.0t4.0t21.0t3, with xin meters and tin
seconds. Find the work done by the force from t0 to t4.0 s.
•••42 Figure 7-41 shows a cord attached to a cart that can slide
along a frictionless horizontal rail aligned along an xaxis. The left
F
:
F
:(cx 3.00x2)i
ˆ
(2.5 x2)i
ˆ NF
:
(x)
along an xaxis from x0 to x9.0 m.The scale of the figure’s verti-
cal axis is set by as6.0 m/s2. How much work has the force done on
the particle when the particle reaches (a) x4.0 m, (b) x7.0 m,
and (c) x9.0 m? What is the particle’s speed and direction of travel
when it reaches (d) x4.0 m,(e) x7.0 m, and (f) x9.0 m?
a (m/s2)
as
0
–as
2468 x (m)
174 CHAPTER 7 KINETIC ENERGY AND WORK
end of the cord is pulled over a pulley, of negligible mass and friction
and at cord height h1.20 m, so the cart slides from x13.00 m to
x21.00 m. During the move, the tension in the cord is a constant
25.0 N. What is the change in the kinetic energy of the cart during
the move?
Module 7-6 Power
•43 A force of 5.0 N acts on a 15 kg body initially at rest.
Compute the work done by the force in (a) the first, (b) the second,
and (c) the third seconds and (d) the instantaneous power due to
the force at the end of the third second.
•44 A skier is pulled by a towrope up a frictionless ski slope
that makes an angle of 12with the horizontal. The rope moves
parallel to the slope with a constant speed of 1.0 m/s. The force
of the rope does 900 J of work on the skier as the skier moves a
distance of 8.0 m up the incline. (a) If the rope moved with a
constant speed of 2.0 m/s, how much work would the force of the
rope do on the skier as the skier moved a distance of 8.0 m up
the incline? At what rate is the force of the rope doing work on
the skier when the rope moves with a speed of (b) 1.0 m/s and
(c) 2.0 m/s?
•45 A 100 kg block is pulled at a constant speed of
5.0 m/s across a horizontal floor by an applied force of 122 N di-
rected 37above the horizontal.What is the rate at which the force
does work on the block?
•46 The loaded cab of an elevator has a mass of 3.0 103kg and
moves 210 m up the shaft in 23 s at constant speed. At what aver-
age rate does the force from the cable do work on the cab?
••47 A machine carries a 4.0 kg package from an initial position
of at t0 to a final posi-
tion of at t12 s. The
constant force applied by the machine on the package is
. For that displacement,
find (a) the work done on the package by the machine’s force and
(b) the average power of the machine’s force on the package.
••48 A 0.30 kg ladle sliding on a horizontal frictionless surface is
attached to one end of a horizontal spring (k500 N/m) whose
other end is fixed. The ladle has a kinetic energy of 10 J as it
passes through its equilibrium position (the point at which the
spring force is zero). (a) At what rate is the spring doing work on
the ladle as the ladle passes through its equilibrium position?
(b) At what rate is the spring doing work on the ladle when the
spring is compressed 0.10 m and the ladle is moving away from the
equilibrium position?
••49 A fully loaded, slow-moving freight elevator has a cab
with a total mass of 1200 kg, which is required to travel upward
54 m in 3.0 min, starting and ending at rest.The elevator’s counter-
weight has a mass of only 950 kg, and so the elevator motor must
help.What average power is required of the force the motor exerts
on the cab via the cable?
••50 (a) At a certain instant, a particle-like object is acted on by a
force while the object’s veloc-
ity is . What is the instantaneous rate
at which the force does work on the object? (b) At some other
time, the velocity consists of only a ycomponent. If the force is un-
changed and the instantaneous power is 12 W, what is the veloc-
ity of the object?
••51 A force acts on a
2.00 kg mobile object that moves from an initial position of
F
:(3.00 N)i
ˆ(7.00 N)j
ˆ(7.00 N)k
ˆ
v
:(2.0 m/s)i
ˆ(4.0 m/s)k
ˆ
F
:(4.0 N)i
ˆ(2.0 N)j
ˆ(9.0 N)k
ˆ
SSM
F
:(2.00 N)i
ˆ(4.00 N)j
ˆ(6.00 N)k
ˆ
d
:
f(7.50 m)i
ˆ(12.0 m)j
ˆ(7.20 m)k
ˆ
d
:
i(0.50 m)i
ˆ(0.75 m)j
ˆ(0.20 m)k
ˆ
ILWSSM
SSM
to a final position of
in 4.00 s. Find (a) the
work done on the object by the force in the 4.00 s interval, (b) the
average power due to the force during that interval, and (c) the an-
gle between vectors and .
•••52 A funny car accelerates from rest through a measured track
distance in time Twith the engine operating at a constant power P.
If the track crew can increase the engine power by a differential
amount dP, what is the change in the time required for the run?
Additional Problems
53 Figure 7-42 shows a cold package of hot dogs sliding right-
ward across a frictionless floor through a distance d20.0 cm
while three forces act on the package. Two of them are horizontal
and have the magnitudes F15.00 N and F21.00 N; the third is
angled down by u60.0and has the magnitude F34.00 N.
(a) For the 20.0 cm displacement, what is the net work done on the
package by the three applied forces, the gravitational force on the
package, and the normal force on the package? (b) If the package
has a mass of 2.0 kg and an initial kinetic energy of 0, what is its
speed at the end of the displacement?
d
:
f
d
:
i
d
:
f(5.00 m)i
ˆ(4.00 m)j
ˆ(7.00 m)k
ˆ
di
:(3.00 m)i
ˆ(2.00 m)j
ˆ(5.00 m)k
ˆ
Figure 7-42 Problem 53.
F
2
F
1
d
F
3
θ
54 The only force acting on a
2.0 kg body as the body moves along
an xaxis varies as shown in Fig. 7-43.
The scale of the figure’s vertical axis
is set by Fs4.0 N. The velocity of
the body at x0 is 4.0 m/s. (a) What
is the kinetic energy of the body at
x3.0 m? (b) At what value of xwill
the body have a kinetic energy of
8.0 J? (c) What is the maximum kinetic energy of the body between
x0 and x5.0 m?
55 A horse pulls a cart with a force of 40 lb at an angle of 30
above the horizontal and moves along at a speed of 6.0 mi/h. (a) How
much work does the force do in 10 min? (b) What is the average
power (in horsepower) of the force?
56 An initially stationary 2.0 kg object accelerates horizontally and
uniformly to a speed of 10 m/s in 3.0 s. (a) In that 3.0 s interval, how
much work is done on the object by the
force accelerating it? What is the instan-
taneous power due to that force (b) at
the end of the interval and (c) at the end
of the first half of the interval?
57 A 230 kg crate hangs from the end
of a rope of length L12.0 m.You push
horizontally on the crate with a
varying force to move it distance d
4.00 m to the side (Fig. 7-44). (a) What is
the magnitude of when the crate is
in this final position? During the crate’s
displacement, what are (b) the total
F
:
F
:
SSM
F
x
(N)
0
–F
s
x(m)
4321
F
s
5
Figure 7-43 Problem 54.
L
d
F
Figure 7-44 Problem 57.
bead for a range of fvalues; W025 J.
How much work is done by if fis (a)
64and (b) 147?
60 A frightened child is restrained by her mother as the child slides
down a frictionless playground slide. If the force on the child from the
mother is 100 N up the slide,the child’s kinetic energy increases by 30 J
as she moves down the slide a distance of 1.8 m. (a) How much work is
done on the child by the gravitational force during the 1.8 m descent?
(b) If the child is not restrained by her mother, how much will the
child’s kinetic energy increase as she comes down the slide that same
distance of 1.8 m?
61 How much work is done by a force ,
with xin meters, that moves a particle from a position
to a position ?
62 A 250 g block is dropped onto a relaxed ver-
tical spring that has a spring constant of k
2.5 N/cm (Fig. 7-46).The block becomes attached to
the spring and compresses the spring 12 cm before
momentarily stopping. While the spring is being
compressed, what work is done on the block by
(a) the gravitational force on it and (b) the spring
force? (c) What is the speed of the block just before
it hits the spring? (Assume that friction is negligi-
ble.) (d) If the speed at impact is doubled, what is
the maximum compression of the spring?
63 To push a 25.0 kg crate up a frictionless
SSM
r
:
f(4 m)i
ˆ(3 m)j
ˆ
(2 m)i
ˆ(3 m)j
ˆr
:
i
F
:(2x N)i
ˆ(3 N)j
ˆ
F
:
a
175
PROBLEMS
work done on it, (c) the work done by the gravitational force on the
crate, and (d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displace-
ment, use the answers to (b), (c), and (d) to find the work your force
does on the crate. (f) Why is the work of your force not equal to
the product of the horizontal displacement and the answer to (a)?
58 To pull a 50 kg crate across a horizontal frictionless floor, a
worker applies a force of 210 N, directed 20above the horizontal.
As the crate moves 3.0 m, what work is done on the crate by (a) the
worker’s force, (b) the gravitational force, and (c) the normal force?
(d) What is the total work?
59 A force is applied to a bead as
the bead is moved along a straight wire
through displacement 5.0 cm. The mag-
nitude of is set at a certain value, but
the angle fbetween and the bead’s
displacement can be chosen. Figure 7-45
gives the work Wdone by on theF
:
a
F
:
a
F
:
a
F
:
a
F
:
65 In Fig. 7-47, a cord runs around
two massless, frictionless pulleys. A
canister with mass m20 kg hangs
from one pulley, and you exert a
force on the free end of the cord.
(a) What must be the magnitude of
if you are to lift the canister at a con-
stant speed? (b) To lift the canister
by 2.0 cm, how far must you pull the
free end of the cord? During that lift,
what is the work done on the canister
by (c) your force (via the cord) and
(d) the gravitational force? (Hint:
When a cord loops around a pulley
as shown, it pulls on the pulley with a
net force that is twice the tension in the cord.)
66 If a car of mass 1200 kg is moving along a highway at
120 km/h, what is the car’s kinetic energy as determined by some-
one standing alongside the highway?
67 A spring with a pointer attached is hanging next to a
scale marked in millimeters. Three different packages are hung
from the spring, in turn, as shown in Fig. 7-48. (a) Which mark on
the scale will the pointer indicate when no package is hung from
the spring? (b) What is the weight Wof the third package?
SSM
F
:
F
:
W (J)
W
0
0
φ
Figure 7-45
Problem 59.
Figure 7-46
Problem 62.
m
F
Figure 7-47 Problem 65.
mm0
30
W
mm0
40
110 N
mm0
60
240 N
Figure 7-48 Problem 67.
68 An iceboat is at rest on a frictionless frozen lake when a sud-
den wind exerts a constant force of 200 N, toward the east, on the
boat. Due to the angle of the sail, the wind causes the boat to
slide in a straight line for a distance of 8.0 m in a direction 20
north of east. What is the kinetic energy of the iceboat at the end
of that 8.0 m?
69 If a ski lift raises 100 passengers averaging 660 N in weight to
a height of 150 m in 60.0 s, at constant speed, what average power
is required of the force making the lift?
70 A force acts on a particle as the particle
goes through displacement . (Other forces
also act on the particle.) What is cif the work done on the particle
by force is (a) 0, (b) 17 J, and (c) 18 J?
71 A constant force of magnitude 10 N makes an angle of 150
(measured counterclockwise) with the positive xdirection as it acts
on a 2.0 kg object moving in an xy plane. How much work is done
on the object by the force as the object moves from the origin to
the point having position vector (2.0 m) (4.0 m) ?j
ˆ
i
ˆ
F
:
d
:(3.0 m)i
ˆ(2.0 m)j
ˆ
F
:(4.0 N)i
ˆcj
ˆ
incline, angled at 25.0to the horizontal, a worker exerts a force of
209 N parallel to the incline. As the crate slides 1.50 m, how much
work is done on the crate by (a) the worker’s applied force, (b) the
gravitational force on the crate, and (c) the normal force exerted
by the incline on the crate? (d) What is the total work done on the
crate?
64 Boxes are transported from one location to another in a ware-
house by means of a conveyor belt that moves with a constant
speed of 0.50 m/s. At a certain location the conveyor belt moves for
2.0 m up an incline that makes an angle of 10with the horizontal,
then for 2.0 m horizontally, and finally for 2.0 m down an incline
that makes an angle of 10with the horizontal.Assume that a 2.0 kg
box rides on the belt without slipping. At what rate is the force of
the conveyor belt doing work on the box as the box moves (a) up
the 10incline, (b) horizontally, and (c) down the 10incline?
176 CHAPTER 7 KINETIC ENERGY AND WORK
72 In Fig. 7-49a, a 2.0 N force is applied to a 4.0 kg block at a
downward angle uas the block moves rightward through 1.0 m
across a frictionless floor. Find an expression for the speed vfof the
block at the end of that distance if the block’s initial velocity is
(a) 0 and (b) 1.0 m/s to the right. (c) The situation in Fig. 7-49bis
similar in that the block is initially moving at 1.0 m/s to the right,
but now the 2.0 N force is directed downward to the left. Find an
expression for the speed vfof the block at the end of the 1.0 m dis-
tance. (d) Graph all three expressions for vfversus downward
angle ufor u0to u90. Interpret the graphs.
rected along the xaxis and has the xcomponent Fax 9x3x2,
with xin meters and Fax in newtons. The case starts at rest at the
position x0, and it moves until it is again at rest. (a) Plot the
work does on the case as a function of x. (b) At what position is
the work maximum, and (c) what is that maximum value? (d) At
what position has the work decreased to zero? (e) At what position
is the case again at rest?
79 A 2.0 kg lunchbox is sent sliding over a frictionless
surface, in the positive direction of an xaxis along the surface.
Beginning at time t0, a steady wind pushes on the lunchbox in the
negative direction of the xaxis. Figure 7-51 shows the position xof
the lunchbox as a function of time tas the wind pushes on the lunch-
box. From the graph, estimate the kinetic energy of the lunchbox at
(a) t1.0 s and (b) t5.0 s. (c) How much work does the force
from the wind do on the lunchbox from t1.0 s to t5.0 s?
SSM
F
:
a
θ
F
θ
F
(
a
)(
b
)
Figure 7-49 Problem 72.
73 A force in the positive direction of an xaxis acts on an object
moving along the axis. If the magnitude of the force is F10ex/2.0
N, with xin meters, find the work done by as the object moves
from x0 to x2.0 m by (a) plotting F(x) and estimating the area
under the curve and (b) integrating to find the work analytically.
74 A particle moves along a straight path through displacement
while force acts on it. (Other
forces also act on the particle.) What is the value of cif the work
done by on the particle is (a) zero, (b) positive, and (c) negative?
75 What is the power of the force required to move a 4500
kg elevator cab with a load of 1800 kg upward at constant speed
3.80 m/s?
76 A 45 kg block of ice slides down a frictionless incline 1.5 m
long and 0.91 m high. A worker pushes up against the ice, parallel
to the incline, so that the block slides down at constant speed.
(a) Find the magnitude of the worker’s force. How much work is
done on the block by (b) the worker’s force, (c) the gravitational
force on the block, (d) the normal force on the block from the sur-
face of the incline, and (e) the net force on the block?
77 As a particle moves along an xaxis,a force in the positive direc-
tion of the axis acts on it. Figure 7-50 shows the magnitude Fof the
force versus position xof the particle.The curve is given by Fa/x2,
with a9.0 Nm2. Find the work done on the particle by the force
as the particle moves from x1.0 m to x3.0 m by (a) estimating
the work from the graph and (b) integrating the force function.
SSM
F
:
F
:(2 N)i
ˆ(4 N)j
ˆ
d
:(8 m)i
ˆcj
ˆ
F
:
F
:
F (N)
12
10
8
6
4
2
0
x
(
m
)
23 4 10
Figure 7-50 Problem 77.
3
2
1
1 3 5 7 8 64
t (s)
2
x(m)
Figure 7-51 Problem 79.
80 Numerical integration. A breadbox is made to move along an
xaxis from x0.15 m to x1.20 m by a force with a magnitude
given by Fexp(2x2), with xin meters and Fin newtons. (Here
exp is the exponential function.) How much work is done on the
breadbox by the force?
81 In the block–spring arrangement of Fig. 7-10, the block’s mass
is 4.00 kg and the spring constant is 500 N/m.The block is released
from position xi0.300 m.What are (a) the block’s speed at x0,
(b) the work done by the spring when the block reaches x0, (c)
the instantaneous power due to the spring at the release point xi,
(d) the instantaneous power at x0, and (e) the block’s position
when the power is maximum?
82 A 4.00 kg block is pulled up a frictionless inclined plane by a
50.0 N force that is parallel to the plane, starting from rest.The nor-
mal force on the block from the plane has magnitude 13.41 N.What
is the block’s speed when its displacement up the ramp is 3.00 m?
83 A spring with a spring constant of 18.0 N/cm has a cage at-
tached to its free end. (a) How much work does the spring force do
on the cage when the spring is stretched from its relaxed length by
7.60 mm? (b) How much additional work is done by the spring force
when the spring is stretched by an additional 7.60 mm?
84 A force N acts on a 2.90 kg
object that moves in time interval 2.10 s from an initial posi-
tion m to a final position r
:
2r
:
1(2.70i
ˆ2.90j
ˆ5.50k
ˆ)
F
:(2.00i
ˆ9.00j
ˆ5.30k
ˆ)
78 A CD case slides along a floor in the positive direction of an
xaxis while an applied force acts on the case. The force is di-F
:
a
m. Find (a) the work done on the object
by the force in that time interval, (b) the average power due to the
force during that time interval, and (c) the angle between vectors
and .
85 At t0, force N begins to act
on a 2.00 kg particle with an initial speed of 4.00 m/s. What is the
particle’s speed when its displacement from the initial point is
m?d
:(2.00i
ˆ2.00j
ˆ7.00k
ˆ)
F
:(5.00i
ˆ5.00j
ˆ4.00k
ˆ)
r
:
2
r
:
1
(4.10i
ˆ3.30j
ˆ5.40k
ˆ)
